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7400
https://www.etymonline.com/word/visionary
Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads. Origin and history of visionary visionary(adj.) 1640s, "of the nature of a vision;" 1650s, "able to see visions;" 1690s, "seen only in visions, unreal;" from vision (n.) + -ary. The sense of "impractical, apt to act on visions as if realities" is attested by 1727. Visional for "pertaining to a vision or visions" is from 1580s. also from 1640s visionary(n.) c. 1700, both as "one able to behold visions" and "one who indulges in impractical fantasies, one who lives in imagination;" from visionary (adj.). A visionist (1660s) is one who believes he sees visions. also from c. 1700 Entries linking to visionary c. 1300, visioun, "that which is seen," specifically "something seen in the imagination or in the supernatural" by one sleeping or waking; from Anglo-French visioun, Old French vision "presence, sight; view, look, appearance; dream, supernatural sight" (12c.), from Latin visionem (nominative visio) "act of seeing, sight, thing seen," noun of action from past-participle stem of videre "to see" (from PIE root weid- "to see"). Also "a narrative account of a vision" (mid-14c.). By early 15c. as "a visual perception" (of something). The meanings "sense of sight, faculty that perceives by the eye;" also "act of seeing external objects" are recorded by late 15c. In 20c. use, "distinct, vivid mental conception of a scheme or anticipation." The meaning "statesman-like foresight, political sagacity" is attested from 1926. adjective and noun word-forming element, in most cases from Latin -arius, -aria, -arium "connected with, pertaining to; the man engaged in," from PIE relational adjective suffix -yo- "of or belonging to." The neuter of the adjectives in Latin also were often used as nouns (solarium "sundial," vivarium, honorarium, cucumerarium "a cucumber field," etc.). It appears in words borrowed from Latin in Middle English. In later borrowings from Latin to French, it became -aire and passed into Middle English as -arie, subsequently -ary. Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads. Trends of visionary More to explore Share visionary Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads. Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads. Want to remove ads? Log in to see fewer ads, and become a Premium Member to remove all ads. Quick and reliable accounts of the origin and history of English words. Scholarly, yet simple. About Support Apps
7401
https://pt.slideshare.net/resala75/hydrostatic-forces-on-plane-surfaces
Uploaded byDr.Risalah A. Mohammed 9,828 views Hydrostatic forces on plane surfaces This document provides a training package on hydrostatic forces on plane surfaces for students in the Environmental Engineering Department. It includes an overview of the topic, objectives, examples, and pre-test and post-test questions. The key ideas covered are how hydrostatic forces form a system of parallel forces on submerged surfaces, how to calculate the magnitude and location of these forces on vertical, inclined, and curved surfaces, and examples demonstrating these calculations. Education◦ In this document Powered by AI Introduction to hydrostatic forces, target audience, aims, and instructional guidelines for students studying fluid mechanics. Outlines performance objectives for understanding hydrostatic forces, includes pre-test questions to assess students' prior knowledge. Overview of hydrostatic forces on plane surfaces, calculation of resulting forces and center of pressure. Various examples illustrating the application of hydrostatic force calculations on plane surfaces. Post-test questions focusing on determining water levels for stability in assessing forces on gates. Answer keys for pre-test and post-test questions on hydrostatic forces, providing solutions to test queries. List of references and textbooks for further reading and study in fluid mechanics and hydrostatics. Downloaded 34 times 1 / 26 2 / 26 3 / 26 4 / 26 5 / 26 6 / 26 7 / 26 8 / 26 9 / 26 10 / 26 11 / 26 Most read 12 / 26 Most read 13 / 26 14 / 26 15 / 26 16 / 26 17 / 26 18 / 26 Most read 19 / 26 20 / 26 21 / 26 22 / 26 23 / 26 24 / 26 25 / 26 26 / 26 More Related Content PDF Fluid Mechanics Chapter 2. 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Mohammed M.Sc. Civil Engineering Asst. Lect. Environmental & Pollution Engineering Department 2011 2. 1- Over view 1-1Target population CH.3: Hydrostatic Forces on Plane Surfaces For the students of second class in Environmental engineering Department in Technical College 3. 1-2 Rationale CH.3: HydrostaticForces on Plane Surfaces The study of force in fluid mechanics is very important in applied field to calculate the amount of force on the walls of the reservoirs, surfaces which exposed to the effects of wind such as aircraft and high buildings. In addition to the application in the field of hydraulic structures to calculate the forces that exposed on dams and gates. 4. 1-3 Central Idea CH.3:Hydrostatic Forces on Plane Surfaces The main goal of this chapter are: 1- Define the hydrostatic force. 2- Applied of the hydrostatic force on surface plane . 5. 1-4 Instructions CH.3: HydrostaticForces on Plane Surfaces 1- Study over view thoroughly 2- Identify the goal of this modular unit 3- Do the Pretest and if you :- Get 9 or more you do not need to proceed Get less than 9 you have to study this modular 4- After studying the text of this modular unit , do the post test and if you :- Get 9 or more , so go on studying modular unit four Get less than 9 , go back and study the modular unit three 6. 1-5 Performance Objectives CH.3:Hydrostatic Forces on Plane Surfaces At the end of this modular unit the student will be able to :- 1- Define the hydrostatic force 2- calculate the magnitude and direction and it position of hydrostatic pressure on vertical, inclined and curved surfaces. 7. 2- Pre test - CH.3:Hydrostatic Forces on Plane Surfaces Q1)) ( 5 mark) The tank in Fig. below is 40cm wide. Compute the hydrostatic forces on horizontal panels BE and AD. Neglect atmospheric pressure . Q2)) ( 5 mark) the net hydrostatic force per unit width on rectangular panel AB in Fig. below and determine its line of action. Not Check your answers in key answer page 8. Hydrostatic Forces onPlane Surfaces CH.3: Hydrostatic Forces on Plane Surfaces  On a plane surface, the hydrostatic forces form a system of parallel forces  For many applications, magnitude and location of application, which is called center of pressure, must be determined.  Atmospheric pressure Patm can be neglected when it acts on both sides of the surface. 9. Hydrostatic Forces onInclined Plane Surfaces CH.3: Hydrostatic Forces on Plane Surfaces The magnitude of FR acting on a plane surface of a completely submerged plate in a homogenous fluid is equal to the product of the pressure PC at the centroid of the surface and the area A of the surface 10. Hydrostatic Forces onInclined Plane Surfaces CH.3: Hydrostatic Forces on Plane Surfaces Line of action of resultant force FR=PCA does not pass through the centroid of the surface. In general, it lies underneath where the pressure is higher.  Vertical location of Center of Pressure is determined by equation the moment of the resultant force to the moment of the distributed pressure force. , xx C p C c I y y y A   11. Example(1) CH.3: Hydrostatic Forceson Plane Surfaces 12. Example(2) CH.3: Hydrostatic Forceson Plane Surfaces 13. Hydrostatic Forces onCurved Surfaces CH.3: Hydrostatic Forces on Plane Surfaces  FR on a curved surface is more involved since it requires integration of the pressure forces that change direction along the surface.  Easiest approach: determine horizontal and vertical components FH and FV separately. 14. Hydrostatic Forces onCurved Surfaces CH.3: Hydrostatic Forces on Plane Surfaces  Horizontal force component on curved surface: FH=Fx. Line of action on vertical plane gives y coordinate of center of pressure on curved surface.  Vertical force component on curved surface: FV=Fy+W, where W is the weight of the liquid in the enclosed block W=rg. X coordinate of the center of pressure is a combination of line of action on horizontal plane (centroid of area) and line of action through volume (centroid of volume).  Magnitude of force FR=(FH 2+FV 2)1/2  Angle of force is a = tan-1(FV/FH) 15. Example(3) CH.3: Hydrostatic Forceson Plane Surfaces 16. Example(3) CH.3: Hydrostatic Forceson Plane Surfaces 17. Example(4) CH.3: Hydrostatic Forceson Plane Surfaces 18. Example(5) CH.3: Hydrostatic Forceson Plane Surfaces 19. Example(6) CH.3: Hydrostatic Forceson Plane Surfaces 20. Example(7) CH.3: Hydrostatic Forceson Plane Surfaces 21. Post test - CH.3: HydrostaticForces on Plane Surfaces Q1)) (5 marks) Gate AB in Fig. below is 16 ft long and 8 ft wide. Neglecting the weight of the gate, compute the water level h for which the gate will start to fall. Q2)) (5 marks) Gate AB in Fig. below a is 16 ft long and 8 ft wide. Neglecting the weight of the gate, compute the water level h for which the gate will start to fall. 22. Answer key CH.3: HydrostaticForces on Plane Surfaces Pretest Q1)) 23. Answer key CH.3: HydrostaticForces on Plane Surfaces Pretest Q2)) 24. Answer key CH.3: HydrostaticForces on Plane Surfaces Post test Q1)) 25. Answer key CH.3: HydrostaticForces on Plane Surfaces Post test Q2)) 26. References CH1: Fluid Properties 1.Evett, J., B. and Liu, C. 1989 “2500 solved problems in fluid mechanics and hydraulics” Library of Congress Cataloging- in-Publication Data, (Schaum's solved problems series) ISBN 0-07-019783-0 2. Rajput, R.,K. 2000 “ A Text Book of Fluid Mechanics and Hydraulic Machines”. S.Chand & Company LTD. 3. White, F., M. 2000 “ Fluid Mechanics”. McGraw-Hill Series in Mechanical Engineering. 4. Wily, S., 1983 “ Fluid Mechanics”. McGraw-Hill Series in Mechanical Engineering.
7402
https://www.youtube.com/watch?v=l1BXOj1TI4Q
1st Derivative Test with Trig - Section 3.3 (Part 2) Mrmathblog 45700 subscribers 9 likes Description 2372 views Posted: 12 Aug 2013 This lesson is very similar to Part 1 of Section 3.3, on the 1st Derivative Test, except we use a trig function here. We find the critical numbers from setting the 1st derivative = 0, then test the intervals of f '(x) to test if it's increasing or decreasing to locate the relative min's and max's. Short, easy lesson. 2 comments Transcript: okay guys uh this one's uh uh part two first derivative test with trigonometry so this should be pretty fast i have one example you're just doing the same thing but with trig okay so if you haven't seen part one see part one to see what i'm talking about so i'm gonna go kind of fast okay so find the intervals on which the function is increasing decreasing uh and then find the relative maximums and minimums of f of x equals one half x minus sine x on the interval zero to two pi all right so to take the derivative first so the derivative of one half x is one half and the derivative of sine is cosine so it's one half x minus cosine so now i set that equal to zero and cosine of x equals a half at pi over three and five pi over three in quadrants 1 and 4. all right so then set up a number line with the endpoints and those criticals these are my criticals right here okay and then i'm going to test a region over here i'll test pi over 4 i'll test a region in here i like testing pi here i'll test seven pi over four and i'm going to test it with my f prime with this guy right here okay so uh at power four it's positive so one minus uh and it's going to be root two over two is going to be a negative okay so i'm going to test those intervals and i'm going to get a negative positive negative when i do that okay because cosine of pi is negative 1 and 1 minus a negative 1 is 1 plus negative 1 so that's positive okay here i'm going to get another negative right here uh uh cosine of pi over 4 and this cosine of 7 power 4 the same root 2 over 2. anyways so that tells me that it's going it's going down and then it starts going back up at power 3 and then it starts going back down at 5 power 3 so there's a relative minimum here there's a relative maximum here and then i plug these values back up into the original right here and i get my relative min to be this value my relative max to be that value and there's where it increases and decreases the intervals i told you this would be a fast one okay so uh if you're in my class there's the homework and there's the answers to to the evens okay nice work you guys
7403
https://www.scribd.com/document/408525986/lesson-plan1-introduction-to-imaginary-and-complex-numbers
Lesson Plan1 - Introduction To Imaginary and Complex Numbers | PDF | Complex Number | Pedagogy Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 184 views 4 pages Lesson Plan1 - Introduction To Imaginary and Complex Numbers This lesson plan introduces students to imaginary numbers. It aims to teach students to define the imaginary unit i, use i to write complex numbers in standard form, and understand that comp… Full description Uploaded by api-457194936 AI-enhanced title and description Go to previous items Go to next items Download Save Save lesson plan1 - introduction to imaginary and compl... For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save lesson plan1 - introduction to imaginary and compl... For Later You are on page 1/ 4 Search Fullscreen 1 Jon Fall INTRODUCTION TO IMAGINARY NUMBERS Lesson Plan for Algebra 2 OVERVIEW & PURPOSE This lesson will introduce the class to the imaginary number i and how we use it to write complex numbers in standard form. EDUCATION STANDARDS 1. HSN.CN.A.1 Know there is a complex number i such that i 2 = -1, and every complex number has the form a + bi with a and b real PRIOR KNOWLEDGE By now, I am assuming the students have already learned how to factor numbers and radicals ESSENTIAL QUESTION 1. What is a complex number? OBJECTIVES 1. TLW be able to define i . 2. TLW use i to put complex numbers in standard form. adDownload to read ad-free 2 MATERIALS ● White Board or projection. ● Drawing sticks for the class ● Textbooks or Chromebooks to access online book VERIFICATION Steps to check for student understanding 1. Warm up activity checks their prior knowledge 2. Warm up review in lesson checks base understanding on imaginary numbers 3. Group collaboration examples check ability to apply their knowledge 4. Answers during the standard form examples helps show how w ell they understand standard form. ACTIVITY ● Warm Up: ○ Factor or Simplify: ■ √9 ■ √32 ■ −2√49 ■ √−64 ○ Answers: ■ 3 ■ 16√2 ■ −14 ■ 8√−1 ○ Review the warm-up by having students volunteer to share their solutions, give guidance only when necessary S-C/S-S ■ Emphasize that in the last problem, we are left with a square root of negative 1. We will be talking about this number a lot today. ● Portfolio Assignment for the Unit: Before we ge t into our math for the day, we need to talk about your unit-long assignment. We are going to be doing a portfolio in this class. Each page of the portfolio should correspon d to a day of class, and you will need to put that day’s homework assignment at the top of each page, just under the date. The following are categories that will be worth the bulk of the points of the portfolio: T-C adDownload to read ad-free 3 ○ Cover page and Table of Contents: The portfolio should have a cover page and an accurate table of contents worth a total of 3 points ○ Summary of the lesson: This segment is worth 3 points per page and should be a summary, in your words, of what was learned that day ○ Solution to challenging problem: This segment is worth 3 points per page and should include your solution to what you thought was the toughest problem and also a personal description of what made it challenging to you ○ Question for next class: This segment is worth 2 points per page and should include a thoughtful question for the next day in class ○ New research and notes for essential questions: This segment will be worth up to 3 points per page and should include notes and thoughts on the essential questions. You can only earn up to 15 points throughout the unit for this section throughout all the pages ○ Corrected problems: After each lesson’s portfolio page, you can include corrections to any and all homework problems that you attempted for full cre dit if solved correctly ● Imaginary Unit: This number, √−1 , is what we call i , the imaginary unit . If a number has a multiple of i in it, that number is imaginary . T-C ○ Which of the numbers in the warm up were real, and which of t hem were imaginary? Talk it through with the class via questioning ○ Lets look at a few more in groups of 3-5, draw sticks to see which group shares their answer with the class G-C/G-G ■ √25 R ■ √−54 I ■ 7i - I ■ 8i 2 – R ● What is i 2 ? Discuss this with the class. ■ i 3 – I ● What is i 3 ? Discuss this with the class ○ Now let’s look at what we did with those last two problems and see what else we can learn. i 2 = -1, i 3 = -i, i 4 = 1, i 5 = i … ● Now we are going to discuss Complex Numbers . Complex numbers make up all Real and Imaginary numbers. They have Real components and Imaginary components. We write the standard form for a complex number as a+bi where a and b are both real numbers and i is our imaginary unit i = √−1 . T-C ○ Examples: Write each number as a complex number in standard form. ■ √49+√−3 6 = 7+6 adDownload to read ad-free 4 ■ √2 5 = 5+0 ■ 2√72+5√−75 = 12√2+25√3 ● The remainder of class should be spent on the assignment: 6-12 even, 13,14,31 ● In the last 5 minutes of c lass, call the class back together and have them share what they learned today or any questions they have with a neighbor, then randomly select a few pairs/groups to tell the class what they shared with each other, including questions if there were any. 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https://www.physoc.org/magazine-articles/reduced-mitochondrial-efficiency-dysfunction-or-defence-in-ageing-muscle/
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Strategy 2023-27 Excellence in physiology News Our partners Our history Board of Trustees, Committees & Groups Governance Staff Diversity and Inclusion Contact us ### Become a member ### Check out our Training Hub Resources for Society Reps Cardiac & Vascular Physiology Epithelia & Membrane Transport Human, Environmental & Exercise Physiology Endocrinology Metabolic Physiology Neuroscience Education & Teaching Mentoring Members reflect on COVID-19 Member support Physiology and COVID-19 Questions from the front line COVID-19 Advisory Panel The Journal of Physiology’s Virtual Journal Club Returning to the lab Engaging with your Theme Conference Attendance Award Education and Teaching Award Research and Knowledge Exchange Award Institutional Engagement Award Paton Historical Studies Fund Unlocking Futures Fund Equity, Diversity and Inclusion Fund Grant for carers Information for Grant recipients Prize Lecture FAQs The Annual Review Prize Lecture The Otto Hutter Teaching Prize and Lecture The Bayliss-Starling Prize Lecture for Mid-Career Physiologists The GL Brown Annual Public Prize Lecture The Hodgkin-Huxley-Katz International Prize Lecture The Joan Mott Prize Lecture Celebrating Women Physiologists The Mabel Fitzgerald Prize for Diversity in Physiology The R Jean Banister Prize Lecture for Early-Career Physiologists The Paton Prize Lecture The President’s Lecture The Sharpey-Schafer Prize Lecture for Translational Physiology The Widening Participation in Physiology Prize Lecture The Rob Clarke Awards Undergraduate Prize for Physiology The Otto Hutter Teaching Prize and Lecture Experimental Physiology Early Career Author Prize Experimental Physiology Mid-Career Researcher Prize Experimental Physiology Inaugural Review Prize The Journal of Physiology Early Investigator Prize Exemplary Service to the Society Award Michael J Rennie Oral Communication Prize Two-Day Scientific Meetings Two-Day Education and Teaching Meetings Webinar series Sandpit Meeting International Development Support Fund Non-Society Meetings Early Career Life Scientists’ Symposium Grant for carers Latest issue Editorial Board of Physiology News Contribute Artificial Intelligence and Health Physiological Considerations for Maximum Indoor Temperatures Climate Change, Physiology, and COP29 Roadmap for Global Heat Resilience Climate and Physiological Resilience Network Developing a Human-centred Heat Resilience Strategy Physiology, Heat and Mental Health Climate Emergency: Research Gaps and Policy Priorities Physiology & Climate Change Translating Knowledge and Research into Impact Knowledge Exchange Case Studies Knowledge Exchange Champions Network Age, health and work Growing Older, Better Supporting the development of public health guidance for long COVID A National Post-Pandemic Resilience Programme Research Excellence Framework (REF) Sport & Exercise Science Education: Impact on the UK Economy Scotland at the Heart of Meeting Global Challenges Physiology and Basic Research Funding in Ireland Contribution of physiology education and training to the UK economy Medical Objectives Physiology and the COVID-19 response Animal research The use of animals in educating the next generation of life science researchers Physiology Passport Age 16–19 Undergraduate and Masters PhD and beyond Reproduction and development Impact of exercise on health Microbiomes in the gut Respiration and lung function Sleep and circadian rhythms Extreme environments Space physiology Conservation of life Excellence in Physiology Award The Physiological Society’s blue plaques Nobel Prize Laureate Members Honorary Fellows Nominate an Honorary Fellow Historical highlights Society timeline Obituaries Oral histories ### What is physiology? ### Meet our members ### Neurophysiological Bases of Human Movement 2025 16 – 17 Dec 2025 Physiology News Magazine Download this issue Home Summer 2006 - Issue Number 63 Editorial Board Contribute Current issue Previous issues Open Drop Past Issues 131-133 Open Drop Issue 133 Issue 132 Issue 131 Past Issues 121-130 Open Drop Issue 130 Issue 129 Issue 128 Issue 127 Issue 126 Issue 125 Issue 124 Issue 123 Issue 122 Issue 121 Past Issues 111-120 Open Drop Issue 120 Issue 119 Issue 118 Issue 117 Issue 116 Issue 115 Issue 114 Issue 113 Issue 112 Issue 111 Past Issues 101-110 Open Drop Issue 110 Issue 109 Issue 108 Issue 107 Issue 106 Issue 105 Issue 104 Issue 103 Issue 102 Issue 101 Past Issues 91-100 Open Drop Issue 100 Issue 99 Issue 98 Issue 97 Issue 96 Issue 95 Issue 94 Issue 93 Issue 92 Issue 91 Past Issues 81-90 Open Drop Issue 90 Issue 89 Issue 88 Issue 87 Issue 86 Issue 85 Issue 84 Issue 83 Issue 82 Issue 81 Past Issues 71-80 Open Drop Issue 80 Issue 79 Issue 78 Issue 77 Issue 76 Issue 75 Issue 74 Issue 73 Issue 72 Issue 71 Past Issues 61-70 Open Drop Issue 70 Issue 69 Issue 68 Issue 67 Issue 66 Issue 65 Issue 64 Issue 63 Issue 62 Issue 61 Past Issues 51-60 Open Drop Issue 60 Issue 59 Issue 58 Issue 57 Issue 56 Issue 55 Issue 54 Issue 53 Issue 52 Issue 51 Past Issues 41-50 Open Drop Issue 50 Issue 49 Issue 48 Issue 47 Issue 46 Issue 45 Issue 44 Issue 43 Issue 42 Issue 41 Past Issues 31-40 Open Drop Issue 40 Issue 39 Issue 38 Issue 37 Issue 36 Issue 35 Issue 34 Issue 33 Issue 32 Issue 31 Past Issues 21-30 Open Drop Issue 30 Issue 29 Issue 28 Issue 27 Issue 26 Issue 25 Issue 24 Issue 23 Issue 22 Issue 21 Past Issues 11-20 Open Drop Issue 20 Issue 19 Issue 18 Issue 17 Issue 16 Issue 15 Issue 14 Issue 13 Issue 12 Issue 11 Past Issues 1-10 Open Drop Issue 10 Issue 9 Issue 8 Issue 7 Issue 6 Issue 5 Issue 4 Issue 3 Issue 2 Issue 1 Summer 2006 - Issue Number 63 Full issue Reduced mitochondrial efficiency: dysfunction or defence in ageing muscle? Mitochondrial oxidative phosphorylation has been found to become less efficient with age in both mouse and human skeletal muscle. Here, David Marcinek explores mechanisms that can lead to either a protective or pathological effect of this uncoupling Features Reduced mitochondrial efficiency: dysfunction or defence in ageing muscle? Mitochondrial oxidative phosphorylation has been found to become less efficient with age in both mouse and human skeletal muscle. Here, David Marcinek explores mechanisms that can lead to either a protective or pathological effect of this uncoupling Features David J Marcinek Department of Radiology, University of Washington Medical Center, Seattle, WA, USA David Marcinek Mitochondria are primary sites for aerobic ATP synthesis and generation of reactive oxygen species (ROS). They also play an important role in regulating cell survival. These multiple roles of mitochondria place them at the centre of the cellular mechanisms responsible for a growing number of pathological conditions. Despite an explosion of research into mitochondrial function in ageing and diseas,e there are important issues that remain unresolved. One of these is the significance of the coupling efficiency of oxidative phosphorylation (P/O). Reduced mitochondrial coupling in ageing tissues presents an interesting paradox. Reduced coupling has been demonstrated to reduce the generation of ROS by the mitochondria. However, uncoupling with age is also associated with impaired ATP synthesis and altered cell energetics. Therefore, certain changes associated with reduced coupling, such as reduced ATP levels, may be detrimental to cell survival, while others, like a decrease in ROS production, may be an adaptive response to cellular stress. Here I focus on mechanisms and significance of reductions in mitochondrial coupling efficiency in ageing muscle. Figure 1. Severe uncoupling in aged mouse skeletal muscle. Mitochondrial O2 consumption and ATP synthesis were determined in vivo in resting mouse hindlimb muscle. The ratio of ATP produced per O2 consumed (P/O) was approximately 50% lower in the old (30-month) mouse muscle than in the young (7-month) mice ( P < 0.05). Data from Marcinek et al. (2005). We have recently demonstrated significant mitochondrial uncoupling in vivo in aged mouse skeletal muscle (Marcinek et al. 2005). Using a combination of magnetic resonance and optical spectroscopy to directly measure ATP use and O2 uptake in vivo, we found that mitochondria in aged mouse skeletal muscle produce on average about 50% fewer ATP molecules per O2 molecules consumed (reduced P/O) than those in young muscle (Fig. 1). This mitochondrial uncoupling is consistent with the reduced in vivo phosphorylation capacity per mitochondrial volume found in the quadriceps from elderly humans (Conley et al. 2000). Mitochondrial ATP synthesis is coupled to oxygen consumption in the mitochondria through the membrane potential generated by proton pumping in the electron transport chain (ETC) (Fig. 2). This membrane potential provides the driving force for protons to flow back into the matrix, which drives ATP synthesis by complex V (F1F0 ATP synthase). Protons may also bypass the ATP synthase and leak across the inner mitochondrial membrane (IMM), short-circuiting the coupling of ATP synthesis and O2 consumption. This leak will dissipate the membrane potential and reduce the efficiency of oxidative phosphorylation (lower P/O). Figure 2. Schematic model of oxidative phosphorylation showing paths for uncoupling of oxidative phosphorylation. Oxygen consumption by the electron transport chain (ETC) and ATP synthesis in the mitochondria are coupled by the membrane potential created by ETC proton pumping across the inner mitochondrial membrane (IMM). The ETC also produces reactive oxygen species (ROS) that cause oxidative stress and can lead to uncoupling. In the case of mild uncoupling, oxidative stress leads to upregulation of proton leak through the uncoupling proteins (UCP). The increased leak reduces ROS generation and decreases the oxidative stress. When oxidative stress exceeds cellular defenses oxidative damage can accumulate causing membrane damage and severe uncoupling. Increased proton leak is the most well accepted mechanism to explain the lower P/O values in aged muscle. This hypothesis is supported by results demonstrating increased proton leak in mitochondria isolated from aged rodent muscle (Harper et al. 2004). Such proton leak may occur as a result of non-specific leak across the IMM resulting from damage to membrane components in aged muscle, or the leak could be mediated through activities of specific membrane proteins, such as the uncoupling proteins (UCP). These leak pathways are not necessarily mutually exclusive, but identifying the roles of membrane damage versus UCP function may be critical to determining whether uncoupling represents mitochondrial dysfunction (damage) or a regulated cellular response to stress (UCP) in ageing muscle. One mechanism for increased mitochondrial proton leak with age is that oxidative damage to mitochondrial proteins and lipids leads to membrane damage making the IMM more permeable to protons. The accumulation of oxidative damage to mitochondria is well documented in aging tissues and damage to membrane lipids has been shown to increase proton leak in isolated systems. Further support for a role for oxidative damage in mitochondrial uncoupling comes from transgenic mice with reduced antioxidant activities. Mitochondria isolated from these mice had higher levels of oxidative damage and lower respiratory control ratios (state 3/state 4 respiration) (Williams et al. 1998), which is indicative of increased proton flux across the IMM. In contrast, an increase in mitochondrial proton leak with age may be due to an upregulation of UCP activity. Increased proton leak through the UCPs would lower the membrane potential and lead to reduced ROS production by the ETC, because mitochondrial ROS production is very sensitive to changes in the resting membrane potential. This means that even a small amount of proton leak due to UCP activity could lead to large decreases in ROS production. This idea has led to the proposal that mitochondrial uncoupling is actively regulated through UCP function as a defence against oxidative damage to mitochondria. Uncoupling by UCP3 in muscle has been found to increase in response to lipid peroxides (Echtay et al. 2003). Since the lipid peroxides form in the mitochondria in the presence of increased oxidative stress their role in upregulating UCP3 activity supports a feedback model where oxidative stress leads to mild uncoupling, which in turn reduces ROS production. The oxidative damage and regulation of UCP hypotheses are not mutually exclusive. This raises the interesting possibility of different degrees of uncoupling in aging tissues – one that may be a regulated cellular defence against oxidative damage, and one that is the result of mitochondrial damage. In the first case mild uncoupling would be due to upregulation of UCP activity in response to oxidative stress. This would create a negative feedback loop by lowering the membrane potential and reducing ROS production. Speakman et al. (2004) found that proton leak was greater in skeletal muscle mitochondria isolated from the longest-lived mice in a population. They demonstrated that UCP3­mediated uncoupling accounted for a significant fraction of the difference between the longer- and shorter-lived individuals. This study provides indirect evidence for a protective effect of mild uncoupling. In contrast, the more severe uncoupling that has been found in aged mouse (Marcinek et al. 2005) and human (Conley et al. 2000) muscles in vivo is more likely associated with mitochondrial damage and dysfunction. The presence of mitochondrial dysfunction is indicated by reductions in the cell’s energy state (ATP/ADP) and/or ATP levels that accompany the lower P/O values in both species. The change in the cellular energetics is a sign that the ability of the mitochondria to meet the energetic demand of the cell is compromised in these tissues. Because reduced ATP levels can be signal for apoptosis, the severe uncoupling may be an early step in the pathway toward tissue degeneration. In the model presented here UCP­mediated mild uncoupling acts as a regulated, adaptive response to reduce mitochondrial oxidative stress to minimize oxidative damage. When mitochondrial oxidative stress increases with age or in disease states, eventually the level of oxidative stress exceeds the ability of the cell to counteract this stress and the result is damage to mitochondrial DNA, proteins, and lipids. This damage accumulates leading to mitochondrial dysfunction, including the severe uncoupling we have identified in vivo in aging skeletal muscle. The associated disruption of cellular energetics and reduced ATP levels may then push the cell toward apoptotic cell death and lead to tissue degeneration. Under this scenario, differences in UCP expression and the degree of mild uncoupling would indicate differences in defense against oxidative damage and contribute to the variation in mitochondrial dysfunction and tissue degeneration found in different tissues. Acknowledgements Thanks to Kevin Conley and Martin Kushmerick for helpful comments during the preparation of this manuscript. Development of these ideas was supported by NIH grants AG00057, AG022385 and AR36281. References Conley KE, Jubrias SA & Esselman PC (2000). Oxidative capacity and ageing in human muscle. J Physiol526, 203-210. Echtay KS, Esteves TC, Pakay JL, Jekabsons MB, Lambert AJ, Portero-Otin M, Pamplona R, Vidal-Puig AJ, Wang S, Roebuck SJ & Brand MD (2003). A signaling role for 4-hydroxy-2-nonenal in regulation of mitochondrial uncoupling. Embo J22, 4103-4110. Harper ME, Bevilacqua L, Hagopian K, Weindruch R & Ramsey JJ (2004). Ageing, oxidative stress, and mitochondrial uncoupling. Acta Physiol Scand182, 321-331. Marcinek DJ, Schenkman KA, Ciesielski WA, Lee D & Conley KE (2005). Reduced mitochondrial coupling in vivo alters cellular energetics in aged mouse skeletal muscle. J Physiol569, 467-473. Speakman JR, Talbot DA, Selman C, Snart S, McLaren JS, Redman P, Krol E, Jackson DM, Johnson MS & Brand MD (2004). Uncoupled and surviving: individual mice with high metabolism have greater mitochondrial uncoupling and live longer. Aging Cell3, 87-95. Williams MD, Van Remmen H, Conrad CC, Huang TT, Epstein CJ & Richardson A (1998). Increased oxidative damage is correlated to altered mitochondrial function in heterozygous manganese superoxide dismutase knockout mice. J Biol Chem273, 28510­28515. 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https://www.infectioncontroltoday.com/view/pseudomonas-aeruginosa-infection-risks-challenges-breakthroughs-for-health-care-professionals
Advanced Technology Advanced Technology Advanced Technology Bug of the Month COVID-19 Environmental Services Environmental Services HAIs Hand Hygiene IC Trends Long-Term Care Long-Term Care Operating Room Personal Protective Equipment Personal Protective Equipment Policy Prevention Prevention Prevention Prevention Sterile Processing Sterile Processing Surface Disinfection Vascular Access News Subscribe Advertisement Pseudomonas aeruginosa: Infection Risks, Challenges, and Breakthroughs for Health Care Professionals By Nameera Temkar Feature Article Pseudomonas aeruginosa, a highly virulent pathogen, poses significant risks to immunocompromised patients, presenting challenges in treatment due to its antibiotic resistance and environmental persistence. Pseudomonas aeruginosa (Adobe Stock 105176221 by Dr_Microbe) Pseudomonas aeruginosa (P aeruginosa) was first described by Charles-Emmanuel Sédillot, who noted that patients' surgical dressings turned blue-green and had a sweet grape-like odor.1 It is a rod-shaped, gram-negative bacteria from the family Pseudomonadaceae. Although it is a facultative aerobe that uses oxygen, it is capable of anaerobic respiration. P aeruginosa can catabolize a wide range of organic molecules for nutrition, allowing it to thrive in various environments such as soil, water, and human skin and mucosa. P aeruginosa has been described as one of the most virulent opportunistic gram-negative bacteria.2 The ability of P aeruginosa to cause a wide variety of infections is due to its large number of virulence factors. Some of them include the following: Apoptosis-inducing cytokines such as ToxA, Azurin, and Pyocyanin Membrane-associated cytokines, such as lipopolysaccharide Type III secretion system exotoxins Infections Caused By Pseudomonas aeruginosa P aeruginosa can thrive in diverse environments such as hot tubs, respirators, and even mops in the hospital, increasing the possibility of exposure and infection.1 As an opportunistic pathogen, it is known to cause mortality in immunocompromised individuals and is the cause of most cases of hospital-acquired pneumonia and respiratory failure. Certain immunocompromised patients, such as those with drug-induced neutropenia, those undergoing chemotherapy, and solid organ transplant recipients, are more vulnerable to P aeruginosa infections. The main reason for this increased risk in solid organ transplant recipients is the use of immunosuppressants. These patients also receive prophylactic antibiotics, which increases the risk of infection with antibiotic-resistant strains. Acute infections Caused By P aeruginosa P aeruginosa can cause serious acute infections such as pneumonia, urinary tract infection, corneal ulcers and keratitis in individuals wearing contact lenses, and bloodstream infections. Pneumonia has been a major burden on health care, with antibiotic-resistant strains further complicating treatment. P aeruginosa has been one of the leading causes of different types of pneumonia. A multinational study of 1173 patients from 54 countries found that P aeruginosa was responsible for 11.3% of community-acquired pneumonia, with 2% of patients suffering from antibiotic-resistant P aeruginosa.3 It was the leading cause of hospital-acquired pneumonia from 1997 to 2008. A recent meta-analysis in Japan showed that P aeruginosa was responsible for 29.2% of ventilator-acquired pneumonia cases, making it responsible for most cases.4 Urinary tract infections caused by P aeruginosa are associated with high morbidity and mortality among elderly patients in the hospital.5 Isolates of P aeruginosa from patients showed higher antibiotic resistance than Escherichia coli, which is the most common cause of urinary tract infections (UTIs). Elderly patients were also seen to have recurrent urinary tract infections UTIs, possibly due to P aeruginosa invading the urinary epithelial cells. It is also a cause of complicated UTIs, especially in patients with catheters, and can lead to life-threatening pyelonephritis.1 Infectious keratitis is a sight-threatening condition common in individuals wearing contact lenses. A multicenter study in Iran found that P aeruginosa was responsible for 71.9% of the cases that tested positive for gram-negative bacteria.6 It has also been associated with severe keratitis.1 Chronic infections caused by P aeruginosa Cystic fibrosis patients are commonly affected by P aeruginosa infections.7 The lungs of patients with cystic fibrosis are a favorable environment for bacterial growth and colonization due to a thick mucus layer, which reduces bacterial internalization, hinders pathogen clearance, and inhibits antimicrobial peptides. A chronic infection with P aeruginosa is generally not responsive to antibiotic therapy, resulting in reduced pulmonary functions and death. P aeruginosa prefers wounds to grow and colonize, and chronic wounds are no exception.1 Chronic wounds such as diabetic foot ulcers, pressure ulcers, and venous leg ulcers are commonly infected by P aeruginosa. Evidence suggests that P aeruginosa may slow wound healing and tissue repair and exacerbate tissue damage. Wounds infected with P aeruginosa were also found to be larger than those infected with other organisms, such as Staphylococcus aureus. Challenges in treating these infections Antibiotic resistance is a significant challenge in all bacterial infections, but more so in infections caused by P aeruginosa. It is resistant to beta-lactamase, quinolones, and aminoglycosides.7 Due to this challenge and the severity of infections caused by this organism, it was placed among other priority pathogens for whom new antibiotics are urgently needed, according to the World Health Organization and the CDC.1 These priority pathogens are also known as ESKAPE (Enterococcus faecium, Staphylococcus aureus, Klebsiella pneumoniae, Acinetobacter baumannii, Pseudomonas aeruginosa, and Enterobacter species). P aeruginosa has 3 types of mechanisms to counter an antibiotic attack—intrinsic resistance, acquired resistance, and adaptive resistance. P aeruginosa possesses a high level of intrinsic resistance to multiple antibiotics.7 Intrinsic resistance refers to a bacteria’s innate ability to reduce the efficacy of an antibiotic due to its structural or functional characteristics. The intrinsic resistance of P aeruginosa includes the expression of effluent pumps that remove antibiotics from the cell, the production of antibiotic-inactivating enzymes, and low other membrane permeability. Acquired resistance of P aeruginosa consists of mutations and horizontal transfer of resistance genes. The formation of biofilm in the lungs of infected patients is used as adaptive resistance. The biofilm works as a diffusion barrier and limits antibiotic access to bacterial cells. Multidrug-resistant persistent cells can form in these biofilms and are responsible for recurrent infections. Latest breakthroughs and strategies Rising rates of antibiotic resistance are a significant burden on health care facilities and the economy. While the development of new therapies has been slow, there is a push to develop new antibiotics and other antibacterial therapies. In recent years, new novel antibiotics with in vitro activity against gram-negative bacteria, including P aeruginosa, have been approved.8 Ceftobiprole, a fifth-generation cephalosporin, has been approved for use in patients with community—and hospital-acquired pneumonia and has shown potent activity against several gram-negative bacteria, such as P aeruginosa and Haemophilus influenza. Ceftolozane-tazobactam, a combination of modified cephalosporin with a beta-lactamase inhibitor, has also shown superiority against other drugs in various trials. A study found that the stimulator of the interferon gene (STING), a pattern recognition receptor, suppressed inflammatory cytokine expression and promoted bacterial killing, reducing the severity of keratitis caused by P aeruginosa.9 Phage therapies against pseudomonas aeruginosa have shown promising results. An advantage of phage therapy over antibiotics is that phage therapy can be used against the biofilm produced by the bacteria.10 A previous study used a cocktail of 4 bacteriophages (Pa193, Pa204, Pa222, and Pa223) against P aeruginosa isolated from patients with chronic rhinosinusitis. All 4 phases alone and in combination were able to significantly decrease the rate of biofilm production after 24 and 48 hours of treatment. It also suggested that using a cocktail increased activity due to a range of hosts and prevented the formation of phage-resistant bacteria. Another study studied the effects of PB1-like, phiKZ-like, and LUZ24-like phages against multidrug-resistance P aeruginosa. LUZ24-like phage was the most effective in destroying the biofilm, which the researchers say is due to its small size. The larger phiKZ-like had the least effect on the biofilm. They also noted that phases may not significantly impact high-density biofilms, but they prevent further accumulation and diffusion of biofilms. P aeruginosa is a highly virulent pathogen that frequently infects immunocompromised individuals. It is also inherently resistant to many antibiotics, making it challenging to treat. While research for novel antibiotics and antibacterial therapies is underway, we urgently need new therapies to be approved for use in patients with multidrug-resistant P aeruginosa. References Wood SJ, Kuzel TM, Shafikhani SH. Pseudomonas aeruginosa: Infections, animal modeling, and therapeutics. Cells. 2023;12(1):199. doi:10.3390/cells12010199 Wood SJ, Goldufsky JW, Seu MY, Dorafshar AH, Shafikhani SH. Pseudomonas aeruginosa cytotoxins: Mechanisms of cytotoxicity and impact on inflammatory responses. Cells. 2023;12(1):195-195. doi:10.3390/cells12010195 Restrepo MI, Babu BL, Reyes LF, et al. Burden and risk factors for Pseudomonas aeruginosa community-acquired pneumonia: A multinational point prevalence study of hospitalised patients. Eur Respir J. 2018;52(2):1701190. doi:10.1183/13993003.01190-2017 Moro H, Aoki N, Matsumoto H, et al. Bacterial profiles detected in ventilator-associated pneumonia in Japan: A systematic review. Respir Investig. 2024;62(3):365-368. doi:10.1016/j.resinv.2024.01.012 Newman J, Floyd R, Fothergill J. Invasion and diversity in Pseudomonas aeruginosa urinary tract infections. J Med Microbiol. 2022;71(3). doi:10.1099/jmm.0.001458 Soleimani M, Tabatabaei SA, Masoumi A, et al. Infectious keratitis: Trends in microbiological and antibiotic sensitivity patterns. Eye. 2021;35(11):3110-3115. doi:10.1038/s41433-020-01378-w Pang Z, Raudonis R, Glick BR, Lin TJ, Cheng Z. Antibiotic resistance in Pseudomonas aeruginosa: Mechanisms and alternative therapeutic strategies. Biotechnol Adv. 2019;37(1):177-192. doi:10.1016/j.biotechadv.2018.11.013 Bassetti M, Magnè F, Giacobbe DR, Bini L, Vena A. New antibiotics for Gram-negative pneumonia. Eur Respir Rev. 2022;31(166). doi:10.1183/16000617.0119-2022 Chen K, Fu Q, Liang S, et al. Stimulator of interferon genes promotes host resistance against Pseudomonas aeruginosa keratitis. Front Immunol. 2018;9. doi:10.3389/fimmu.2018.01225 Chegini Z, Khoshbayan A, Taati Moghadam M, Farahani I, Jazireian P, Shariati A. Bacteriophage therapy against Pseudomonas aeruginosa biofilms: A review. Ann Clin Microbiol Antimicrob. 2020;19(1). doi:10.1186/s12941-020-00389-5 Newsletter Stay prepared and protected with Infection Control Today's newsletter, delivering essential updates, best practices, and expert insights for infection preventionists. Subscribe Now! Recent Videos Top 5 Infection Prevention Articles of Summer 2025 Bug of the Month: I Like to Hitch a Ride Evidence, Trust, and Prevention: David J. 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7406
https://www.khanacademy.org/science/class-11-chemistry-india/xfbb6cb8fc2bd00c8:in-in-thermodynamics/xfbb6cb8fc2bd00c8:in-in-measurement-of-change-in-u-and-h-calorimetry/v/worked-example-enthalpy-and-calorimetry
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7407
https://math.stackexchange.com/questions/2588979/write-lines-as-affine-combinations-of-points
projective geometry - Write lines as affine combinations of points - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Write lines as affine combinations of points Ask Question Asked 7 years, 9 months ago Modified7 years, 9 months ago Viewed 680 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I'm learning projective geometry and need help with the following exercise : We define the lines l 1 l 1 and l 2 l 2 in A 2={z=1}⊂R 3∖{0}A 2={z=1}⊂R 3∖{0} as the intersection of A 2 A 2 with the planes x−y=1 x−y=1 and x+z=2 x+z=2 respectively. Find those lines and write them as affine combinations of points. Since I'm having difficulties for the second part of the question I'm going to share my thoughts on the first part. When we embed A 2 A 2 into R 3 R 3 as the plane z=1 z=1, we tacitly add the latter equation to whatever planar equations we're working with. Therefore, the line given by the equation x−y=1 x−y=1 in A 2 A 2 becomes the system of equations {x−y z=1=1 in R 3.{x−y=1 z=1 in R 3. Similarly, the line given by the equation x+z=2 x+z=2 in A 2 A 2 becomes the system of equations {x+z z=2=1 in R 3.{x+z=2 z=1 in R 3. I'm not sure what the author means by "Find those lines" but this is my answer to the first part of the question. Is it correct and/or do I need to add anything else? For the second part of the question I have no idea. I know the definition of an affine combination but I don't know how to write the given lines as affine combinations of points. Any help would be much appreciated. projective-geometry projective-space affine-geometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Jan 2, 2018 at 15:55 user347616 user347616 3 Hint: What geometric object does the set of affine combinations of two given points describe?celtschk –celtschk 2018-01-02 15:58:50 +00:00 Commented Jan 2, 2018 at 15:58 @celtschk They describe a line. Thank you for the hint. Is is still unclear to me how I should continue from here. For instance, the points (1,0,1)(1,0,1) and (2,1,1)(2,1,1) both belong to the line l 1 l 1. So there affine combination describe the line l 1 l 1. Is that correct? It is also unclear to me how I should properly write this affine combination.user347616 –user347616 2018-01-02 22:09:30 +00:00 Commented Jan 2, 2018 at 22:09 It is correct, the affine combinations describe the line through the two points, that is, l 1 l 1. I would write the set of affine combinations of the two points a a and b b as {λ a+(1−λ)b|λ∈R}{λ a+(1−λ)b|λ∈R}.celtschk –celtschk 2018-01-02 23:17:46 +00:00 Commented Jan 2, 2018 at 23:17 Add a comment| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. “Find those lines” means solve the systems of equations that you’ve set up to get a pair of points p p and q q in the intersection. You can then express the line as (1−λ)p+λ q(1−λ)p+λ q, λ∈R λ∈R. In general, an affine combination is a linear combination for which the sum of the coefficients is 1 1. Here, this serves to keep the resulting point on the z=1 z=1 plane. On the projective plane, an affine combination isn’t enough to capture all of the points on a line. If both p p and q q are finite, (1−λ)p+λ q(1−λ)p+λ q misses the line’s point at infinity. To allow for this, you must instead use the set of linear combinations λ p+μ q λ p+μ q, with λ λ and μ μ not both zero—the join of the two points. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 3, 2018 at 12:04 answered Jan 3, 2018 at 8:49 amdamd 55.2k 3 3 gold badges 40 40 silver badges 100 100 bronze badges 1 I remember you helping me on the related subject of the intersection of lines in the projective plane.Your answer was once again extremely helpful. As a bonus we can find the relation between the two given lines. They intersect in A 2 A 2 at a point, say P P, with homogeneous coordinates [1:−1:−1]×[1:0:−1]=[1:0:1][1:−1:−1]×[1:0:−1]=[1:0:1].user347616 –user347616 2018-01-03 11:32:04 +00:00 Commented Jan 3, 2018 at 11:32 Add a comment| You must log in to answer this question. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 62Lines in projective space 11Conic by three points and two tangent lines 1The projective space of all lines through the origin 1Lines in dual projective space 2Intersection of lines in a subspace of projective 3 3-space 2Projective line and homogeneous coordinates 2Point of intersection of lines in R P 2 R P 2 0How can affine plane extended of projective plane? 3Describing affine subspace Hot Network Questions Direct train from Rotterdam to Lille Europe How to start explorer with C: drive selected and shown in folder list? How to locate a leak in an irrigation system? Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? How do you emphasize the verb "to be" with do/does? 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https://www.youtube.com/watch?v=p7cYtUyYFTY
How to Graph Linear Equations | Graph y = x Midnight Math Tutor 6910 subscribers 37 likes Description 4549 views Posted: 3 Apr 2023 They day has arrived! We are going to look at an equation in slope-intercept form and graph it! We can do this pretty easily by identifying the slope and y-intercept. Plot the y-intercept and then count to the next point by using the slope. Watch the second example for a special case when the equation is just y = x. midnightmathtutor #graphaline #linearequations Enjoying our videos? Give us a like and subscribe! Fonts used in thumbnail/logo: KG Always a Good Time, KG Drops of Jupiter 10 comments Transcript: hey guys it's a big exciting day we're going to graph some lines from these equations okay you've probably been building up to this so basically when you first started this unit or chapter or whatever you were probably plugging in numbers for x and getting a y that was your ordered pair and then you were um plotting those and making a line right that is great but guess what this is a faster way we like faster right get your homework done sooner so what we're going to do is we are going to look for what we call slope intercept form okay if our equation is set up like this the information we need is right in front of our face okay so basically if you're asked to graph a line if you can get it into this form where you have y equals if this is the m will be a number okay a number X plus or minus another number okay if you're like what are you talking about just hang with me so you see this here we have y equals negative 2x minus seven this is in slope intercept form because Y is by itself okay if y weren't by itself I could add or subtract or do whatever I needed to do to both sides to get it in this form okay now when it is in this form the number in front of my X is my slope okay this is my slope okay the number being added or subtracted on the back is my y-intercept now you're probably familiar with these terms but if you're not stick around we're going to do it together okay so what this tells me is my line for this equation crosses my y-axis that's oh sorry that's this one right crosses my y-axis at negative seven okay so it's going to look something like this star 0 go down seven one two three four five six seven right there okay I know my line crosses that point okay but I don't know if it goes like this like this how steep it is what tells me that is my slope okay so my slope is negative two okay now we like to write slope as a fraction okay so I'm going to write it as negative 2 you're like that's not a fraction if a number is not a fraction we can always turn it into one by just putting it over a 1. okay because negative 2 divided by 1 is negative 2. so it's still just negative two okay when we have slope um kind of a easy way to remember it is rise over run you've probably heard that before and basically all that means is you do up and down over right and left okay so what that means is my slope is negative two over one so that means I'm going to do up and down first so I'm starting here my slope is negative two so I'm going to go down two over one it's a positive one so I'm going to go to the right one okay and there is my next Point okay I could go down two over one again it's off my graph but it would be about there right and then it's kind of helpful to find some that are that way as well so negative two over one can also be written as 2 over negative one okay because when I divide that it's going to simplify down to just negative 2 again okay so I can also think of it as starting of this point that my up and down or my rise is 2 and negative 1 would go left one okay hopefully that makes sense so up to left one and I could keep going for literally ever okay but I'm just gonna do a few more okay there we go so there is my line well it's not a line yet it's just dots hold on here is my line and it literally goes on forever okay so that means all the ordered pairs along this line are answers for this equation okay all right let's do one that looks a little funny Y equals X okay it's like what does that even mean so we're in slope intercept form right because Y is by itself but where's my M and where's my B okay well the m is pretty easy to find because this is really just One X right we just don't usually write the one but it really is One X right so M or my slope is one okay so I'm gonna write that down my slope or a lot of times now you'll see it written as m my slope is one again we like slope to be a fraction so how do we make this into a fraction it's just over one right because 1 divided by 1 is 1. okay and then what about my y-intercept well hang with me I could write this as plus zero right that doesn't change what my equation is because anything plus 0 is just itself so that means my y-intercept is zero okay there we go so now let's graph this okay so I'm use orange we start with our y-intercept my y-intercept is zero so that means it crosses this line at zero which is right here okay my slope is one over one so since they're both positive I'm going to go up and right okay so up one over one up one over one up one over one okay and I could keep going right now I could also write this as negative 1 over negative one okay does that sound crazy so this could also be written as negative one over negative one y because negative 1 divided by negative 1 simplifies down to one so it's the same thing okay basically all that is saying is I can go this way too down one left one down one left one okay and again I could literally go forever okay so there is my line okay now if you run into one that just looks like y equals 4 or x equals one I will link a video with some examples of those ones okay all right but hopefully that's made sense and good luck getting your homework done bye
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https://mmerevise.co.uk/gcse-maths-revision/edexcel-igcse/arithmetic-sequences-and-sums/
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This rule allows us to find a value at the n th position in the sequence, which we call the n th term. We are also able to use the n th term rule to take asum of the first n terms of a sequence. Finding the n th term An arithmetic sequence is a sequence where every term has the same difference d, between it and the next term. We can use the n th term rule to generate any value in an arithmetic sequence. The rule is The n th term = a + (n-1)d , where d is the common difference, and a is the very first term in the sequence. Example: Find the 10 th term in the following sequence, Step 1: Find the first term (a) In this sequence the first term is given as 3. Step 2: Find the common difference (d) The common difference is the difference between two consecutive terms. In this case we can chose the first and second terms to find d, d = 2\text{nd term} - 1\text{st term} = 7 - 3 = 4. Step 3: Use the formula a = 3, d = 4, so the 10 th term is 3 + (10 - 1) \times 4 = 3 + 9\times 4 = 3 + 36 = 39 So the 10 th term in the sequence is 39. Level 6-7 GCSE Edexcel iGCSE Finding the Sum of a Sequence We can use a formula to find thesum of a sequencefrom the first term to the n th term, where n is the number of terms. We denote this sum as S_n There are two formulas that can be used, although they are essentially the same. The first formula: S_n = \dfrac{n}{2} (2a + (n-1)d), Where a is the first term in the sequence, n is thenumber of terms and d is the common difference. From this first formula can find the second formula. We can deduce that 2a + (n-1)d = a + (a + (n-1)d) and by the n th term rule we know that the n th term = a + (n-1)d. We can call the n th term a_n and the first term a = a_1. So 2a + (n-1)d = a_1 + a_n, which can be substituted back into the first formula. The second formula: S_n = \dfrac{n}{2} (a_1 + a_n) Where a_1 is the first term,a_n is the final term, and n is the number of terms. Level 6-7 GCSE Edexcel iGCSE Example 1: Finding the 13 th Term Consider the sequence: 1, 4, 7, 10, \dots Find the 13 th term in the sequence. [3 marks] Firstly we can find thecommon difference: 2 nd term - 1 st term = 4 - 1 = 3 3 nd term - 2 st term = 7 - 4 = 3 4 nd term - 3 st term = 10 - 7 = 3 The common difference (d) = 3. The first term (a) =1 . This gives us the n th term rule of a_n = 1 + (n-1)\times 3. For the 13 th term, a_{13} = 1 + (13-1) \times 3 a_{13} = 1 + 12 \times 3, a_{13} = 1 + 36, a_{13} = 37, The 13 th term of the sequence is 37. Level 6-7 GCSE Edexcel iGCSE Example 2: Finding the sum to the 1000 th term Consider the sequence: 1, 2, 3, 4, \dots Find the sum of the first 1000 terms in the sequence. [3 marks] The common difference, d = 1. The first term, a = 1 We can use the first formula for sum of an arithmetic sequence to find the sum up to n =1000. S_{1000} = \dfrac{1000}{2} (2 \times 1 + (1000-1) \times 1) S_{1000} = \dfrac{1000}{2} (2 + (999)) S_{1000} = \dfrac{1000}{2} (1001) S_{1000} = 500 \times (1001) S_{1000} = 500,500 Level 6-7 GCSE Edexcel iGCSE Example 3:Determining if a value is a term in a sequence Consider the sequence: 15, 21, 27, 33, \dots determine whether or not the value 99 is a term in the sequence. [3 marks] Firstly we must find the common difference: 2 nd term - 1 st term = 21 - 15 = 6 3 rd term - 2 nd term = 27 - 21 = 6 4 th term - 3 rd term = 33 - 27 = 6 The common difference (d) = 6. Therefore then th term ruleis 15 + (n - 1) \times 6 Now we can equate that to 99 to see if it is a term, 99 = 15 + (n-1) \times 6, 84 = (n-1) \times 6, 84 \div 6 = n - 1, 14 = n - 1, n = 15, So yes, 99 is the 15 th term in the sequence. Level 6-7 GCSE Edexcel iGCSE Example 4: Using the Second Summation Formula An arithmetic sequence has two known terms. The first term is a_1 = 7 and the 43 rd term is a_{43} = 343. Find the n th term rule and the sum of the sequence up to the 43 rd term. [4 marks] Then th term rule: a_n = a_1 + (n-1)d Let n = 43 then, a_{43} = 343 = 7 + (43 - 1)d, 343 - 7 = 42d, 336 = 42d, 336 \div 42 = d, d = 8. The n th term rule then is: a_n = 7 + (n-1)\times 8 The sum to the 43 rd term is: S_{43} = \dfrac{43}{2} \times (a_1 + a_{43}) S_{43} = \dfrac{43}{2} \times (7 + 343) S_{43} = \dfrac{43}{2} \times 350 S_{43} = 7525 Level 6-7 GCSE Edexcel iGCSE Arithmetic Sequences and Sums Example Questions Question 1:The first 5 terms of an arithmetic sequence are 3,\,\,8,\,\,13,\,\,18,\,\,23 Find the formula for the n^{th} term of this sequence. [2 marks] Level 6-7 GCSE Edexcel iGCSE The first term, a = 3. We must now find the common difference. 2 nd term - 1 st term = 8 - 3 = 5. The common difference, d = 5. So the n th term formula is 3 + (n-1) \times 5=5n-2 Save your answers with Gold Standard Education Sign Up Now Show Answer Still marking manually? Save your answers with Find out more Question 2:The first 5 terms of an arithmetic sequence are -6,\,\,-4,\,\,-2,\,\,0,\,\,2 Find the sum of the first 30 terms of this sequence. [4 marks] Level 6-7 GCSE Edexcel iGCSE The first term, a = -6. We must now find the common difference. 2 nd term - 1 st term = -4 - (-6) = 2 . The common difference, d = 2. Summation formula: S_n = \dfrac{n}{2} (2a + (n-1)d) , so, S_{30} = \dfrac{30}{2} (2 \times (-6) + (30 - 1) \times 2) S_{30} = 15 (-12 + (29) \times 2) S_{30} = 15 (-12 + 58) = 15 \times 46 = 690 Save your answers with Gold Standard Education Sign Up Now Show Answer Still marking manually? Save your answers with Find out more Question 2: A sequence is defined by the formula 1080 + (n-1)(-40) a) Work out the first 5 terms of this sequence. [2 marks] b) Determine whether or not -140 is in the sequence. [2 marks] Level 6-7 GCSE Edexcel iGCSE a) To generate the first 5 terms of this sequence, we will substitute n=1, 2, 3, 4, 5 into the formula given. 1 st = 1080 + ((1)-1)(-40) = 1080 2 nd = 1080 + ((2)-1)(-40) = 1040 3 rd = 1080 + ((3)-1)(-40) = 1000 4 th = 1080 + ((4)-1)(-40) = 960 5 th = 1080 + ((5)-1)(-40) = 920 So, the first 5 terms are 1080, 1040, 1000, 960, and 920 b) We can equate -140 to the n th term rule, to find it’s position in the sequence. If it’s position is a whole number, it must be in the sequence, if not it can’t be. -140 = 1080 + (n-1)(-40) -1220 = (n-1)(-40) -1220 \div -40 = (n-1) -1220 \div (-40) = 30.5 = (n-1) n = 31.5 As n is not an integer, -140 is not in the sequence. Save your answers with Gold Standard Education Sign Up Now Show Answer Still marking manually? Save your answers with Find out more Specification Points Covered Edexcel iGCSE Sequences, Functions and Graphs 3.1A – Understand and use common difference (d) and first term (a) in an arithmetic sequence. 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https://brainly.com/question/10371453
[FREE] Identify the solution of the recurrence relation a_n = 6a_{n-1} - 8a_{n-2} for n \geq 2, together with the - brainly.com 5 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +30,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +45k Ace exams faster, with practice that adapts to you Practice Worksheets +6,2k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Identify the solution of the recurrence relation a n​=6 a n−1​−8 a n−2​ for n≥2, together with the initial conditions a 0​=4, a 1​=10. 2 See answers Explain with Learning Companion NEW Asked by rutu7404 • 06/04/2018 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 101884153 people 101M 5.0 0 Upload your school material for a more relevant answer Via the generating function method, let G(x)=n≥0∑​a n​x n Then take the recurrence, a n​=6 a n−1​−8 a n−2​ multiply everything by x n and sum over all n≥2: n≥2∑​a n​x n=6 n≥2∑​a n−1​x n−8 n≥2∑​a n−2​x n Re-index the sums or add/remove terms as needed in order to be able to express them in terms of G(x): n≥2∑​a n​x n=n≥0∑​a n​x n−(a 0​−a 1​x)=G(x)−4−10 x n≥2∑​a n−1​x n=n≥1∑​a n​x n+1=x n≥1∑​a n​x n=x(G(x)−a 0​)=x(G(x)−4) n≥2∑​a n−2​x n=n≥0∑​a n​x n+2=x 2 n≥0∑​a n​x n=x 2 G(x) So the recurrence relation is transformed to G(x)−4−10 x=6 x(G(x)−4)−8 x 2 G(x) (1−6 x+8 x 2)G(x)=4−14 x G(x)=1−6 x+8 x 2 4−14 x​=(1−4 x)(1−2 x)4−14 x​=1−4 x 1​+1−2 x 3​ For appropriate values of x, we can express the RHS in terms of geometric power series: G(x)=n≥0∑​(4 x)n+3 n≥0∑​(2 x)n=n≥0∑​(4 n+3⋅2 n)x n which tells us that a n​=4 n+3⋅2 n Answered by LammettHash •5.9K answers•101.9M people helped Thanks 0 5.0 (4 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 101884153 people 101M 5.0 0 Introduction to Groups, Invariants and Particles - Frank W. K. Firk Quantum Mechanics in Chemistry - Jack Simons Quantum Mechanics III - Y. D. Chong Upload your school material for a more relevant answer The solution to the recurrence relation a n​=6 a n−1​−8 a n−2​ with initial conditions a 0​=4 and a 1​=10 is a n​=3(2 n)+4 n. This is derived from the characteristic polynomial and solving for constants using the initial conditions. The constants are found to be A=3 and B=1. Explanation To solve the recurrence relation a n​=6 a n−1​−8 a n−2​ with initial conditions a 0​=4 and a 1​=10, we begin by determining the characteristic equation associated with the recurrence relation. This characteristic equation is found by substituting a n​=r n into the recurrence, which gives us: r n=6 r n−1−8 r n−2 Dividing by r n−2 (for r=0) yields: r 2−6 r+8=0 We can now factor this quadratic equation as: (r−2)(r−4)=0 Therefore, the roots are r 1​=2 and r 2​=4. The general solution for a recurrence relation with distinct roots is: a n​=A(2 n)+B(4 n) where A and B are constants determined by the initial conditions. To find these constants, we will substitute the initial conditions into the general solution. For n=0: a 0​=A(2 0)+B(4 0)=A+B=4 For n=1: a 1​=A(2 1)+B(4 1)=2 A+4 B=10 Now, we have a system of equations: 1. A+B=4 2. 2 A+4 B=10 Let's solve these equations. From the first equation, we can express A in terms of B: A=4−B Substituting this into the second equation: 2(4−B)+4 B=10 8−2 B+4 B=10 2 B=2 B=1 Now substituting B=1 back into the equation for A: A=4−1=3 Therefore, we have the constants: A=3 and B=1. Putting it all together, the particular solution to our recurrence relation is: a n​=3(2 n)+1(4 n) Thus, the final solution is: a n​=3×2 n+4 n This expression will provide the value of a n​ for any n≥0. Examples & Evidence An example would be to find a 2​. By substituting n=2 into the solution, we get: a 2​=3(2 2)+4 2=3(4)+16=12+16=28. The calculation of the characteristic polynomial and its roots follows established methods in solving linear recurrence relations, indicating that the solution lies in the form provided. Thanks 0 5.0 (4 votes) Advertisement Community Answer This answer helped 5244089 people 5M 5.0 0 The solution to the given recurrence relation on the conditions a0 = 4 and a1 = 10, is an = 2(2)^n + 2(4)^n. This was found using the roots of the characteristic equation and the given initial conditions. Explanation The problem at hand involves a recurrence relation, which is a form of mathematical equation used often in computer science and discrete mathematics. The given recurrence relation is: an = 6an−1 - 8an−2, with the initial conditions a0 = 4 and a1 = 10. To solve this we begin with the characteristic equation of the recurrence relation which is r² - 6r + 8 = 0. The roots of the characteristic equation give us the general solution for the recurrence relation so we solve for 'r'. Our roots are r = 2 and r = 4. The general solution given the roots can be expressed as an = C(2)^n + D(4)^n. To find the constants C and D, we use the given initial conditions. Plugging in a0 = 4 gives us C + D = 4 and plugging in a1 = 10 gives us 2C + 4D = 10. Solving this system of equations, we find C = 2 and D = 2. Therefore, the solution to the recurrence relation an = 6an−1 - 8an−2, with initial conditions a0 = 4 and a1 = 10, is an = 2(2)^n + 2(4)^n. Learn more about Recurrence Relation here: brainly.com/question/34380129 SPJ3 Answered by CharlesBronson •13.2K answers•5.2M people helped Thanks 0 5.0 (1 vote) Advertisement ### Free Mathematics solutions and answers Community Answer Solve these recurrence relations together with the initial conditions given. an = an−1 + 6an−2 for n ≥ 2, a0 = 3, a1 = 6 Community Answer 5.0 Solve this recurrence relation together with the initial condition given. an = 2an−1 + 5an−2 − 6an−3 with a0 = 7, a1 = −4, and a2 = 8 Community Answer solve the following recurrence relations (represent an in term of n alone) a0 = 1, a1 = 3, an = 6an−1 − 8an−2 for n ≥ 2 Community Answer Find the first five terms of the sequence defined by each of these recurrence relations and initial conditions. a) an = 6an-1, a0 = 2 b) an = −2an-1, a0 = −1 c) an = an-1 – an-2, a0 = 2, a1 = −1 Community Answer 4. (10+10 points) (a) Solve the recurrence relation an = 4an−1 + 4an−2 with initial conditions a0 = 1 and a1 = 2. (b) Find all solutions of the recurrence relation an = 2an−1 + 3 · 2n. Community Answer Solve these recurrence relations together with the initial conditions given. a) an = an−1 + 6an−2 for n ≥ 2, a0 = 3, a1 = 6 b) an = 7an−1 − 10an−2 for n ≥ 2, a0 = 2, a1 = 1 c) an = 6an−1 − 8an−2 for n ≥ 2, a0 = 4, a1 = 10 d) an = 2an−1 − an−2 for n ≥ 2, a0 = 4, a1 = 1 e) an = an−2 for n ≥ 2, a0 = 5, a1 = −1 f ) an = −6an−1 − 9an−2 for n ≥ 2, a0 = 3, a1 = −3 g) an+2 = −4an+1 + 5an for n ≥ 0, a0 = 2, a1 = 8 Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? New questions in Mathematics Factor the expression 3 5​m+2. Use 3 1​ as the common factor. Write an equation of the line that passes through (−11,−10) and is parallel to the y-axis. 60 minutes from now, the time will be 4:50 PM. What time was it 90 minutes ago? Solve the following equation. Separate multiple answers with a comma. If there is no solution, write NS for your answer. 5 x=4 x−3​+4 Your start-up company has invented a new water filter. If the filter can process 10 gallons of water per minute, how many gallons of water can it process in one 24-hour day? 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DESPEJE DE FORMULA DE CAÍDA LIBRE - VARIABLE TIEMPO (t) iEnciclotareas 411000 subscribers 48 likes Description 3188 views Posted: 29 Sep 2021 DESPEJE DE FÓRMULA DE CAÍDA LIBRE - VARIABLE TIEMPO (t) En este video despejamos la variable tiempo (t) de una de las fórmulas de altura del movimiento de caída libre. Formula: h= ½gt^2. Donde; h= altura, g= aceleración de la gravedad t= tiempo despeje #formula #tiempo Por favor comparte este video y coméntales a tus amigos sobre este canal. Ayúdanos a crecer, saludos cordiales y bendiciones. SUSCRÍBETE, ES GRATIS No olvides compartir y darle a me gusta. Si tienes alguna duda comenta. Activa la 🔔Campanita - NOTIFICACIONES. ¡Sígueme! Youtube: Facebook: Instagram: Twitter: iEnciclotareas- Canal de Ejercicios Resueltos de Ingeniería Civil Por Angel Suarez. Ingeniero Civiil. Rep.Dom. Música de fondo: Stoker, de la biblioteca de YouTube 2 comments Transcript: hola qué tal te habla el ingeniero ángel suárez este vídeo trata acerca de después de variables en esta ocasión vamos a trabajar con la fórmula o con una de la fórmula de caída libre el enunciado dice lo siguiente después de la siguiente fórmula el tiempo aquí tenemos la fórmula que nos dice que altura es igual a un medio multiplicado por la aceleración de la gravedad multiplicada por la variable tiempo elevado al cuadrado de esta fórmula vamos a despejar la variable tiempo y lo que hacemos es que tomamos la fórmula altura es igual a un medio multiplicado por la aceleración de la gravedad multiplicada por el tiempo al cuadrado lo primero que hago es acomodar la fórmula ya que un medio multiplicado por aceleración de la gravedad multiplicada por el tiempo al cuadrado es lo mismo que decir aceleración a la gravedad por el tiempo al cuadrado dividido entre 2 esta fórmula que ustedes vienen aquí es lo mismo que esta fórmula que tenemos aquí entonces de esta fórmula vamos a despejar la variable tiempo lo primero que hago es trabajar con el 2 que tengo aquí que como está dividiendo si yo lo quiero eliminar de aquí debo de multiplicarlo tanto a la izquierda como a la derecha x 2 por lo tanto voy a tener lo siguiente 2 x altura es igual al producto de la aceleración de la gravedad por el tiempo al cuadrado dividido entre 2 x 2 aquí yo tacho 2 con 2 y nos queda que todos x altura es igual a aceleración de la gravedad multiplicada por el tiempo al cuadrado es decir que ahora la fórmula es la siguiente 2 x altura es igual al producto de la aceleración de la gravedad por el tiempo al cuadrado como quiero despejar la variable tiempo voy a trabajar con la variable aceleración de la gravedad que como está multiplicando aquí si yo la quiero eliminar de aquí debo de dividir tanto lo que está a la izquierda como lo que está a la derecha del signo de igual entre la variable aceleración de la gravedad quiere decir que ahora la fórmula es la siguiente 2 x altura dividido entre aceleración de la gravedad es igual a aceleración de la gravedad multiplicada por el tiempo al cuadrado dividido entre aceleración de la gravedad aquí yo tacho aceleración de la gravedad con la aceleración de la gravedad y nos queda que 2 x altura dividido entre aceleración de la gravedad es igual al tiempo al cuadrado es decir que ahora la fórmula es la siguiente 2 x altura dividido entre aceleración de la gravedad es igual al tiempo al cuadrado que es lo mismo decir tiempo al cuadrado es igual a 2 x altura dividido entre aceleración de la gravedad pero aquí yo tengo el tiempo al cuadrado y el enunciado dice que despeje el tiempo por lo tanto yo debo de eliminar este exponente cuadrado de aquí para lo cual voy a colocar raíz cuadrada a ambos lados tanto a la izquierda como a la derecha de siendo igual yo sé que la raíz cuadrada del tiempo al cuadrado es tiempo por lo tanto voy a tener que tiempo es igual a la raíz cuadrada de 2 x la altura dividido entre aceleración de la gravedad esta es la variable tiempo despejada es decir que de la fórmula que me dice que altura es igual a un medio x aceleración de la gravedad multiplicada por el tiempo elevado al cuadrado si yo despejo la variable tiempo voy a tener que tiempo es igual es cuadrada de 2 x la altura dividido entre aceleración de la gravedad y bien hasta aquí es nuestro vídeo acerca de después de variables en esta ocasión trabajamos con una de la fórmula del movimiento de caída libre donde despejamos la variable tiempo y bien si te gustó el vídeo de la me gusta si tienes alguna duda comenta ley o te la responderé no olvides compartir este vídeo y suscribirte recuerda que subimos vídeos de análisis estructural física matemática y otras bendiciones y hasta la próxima
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Equations in the Form px + q = r and p(x + q) = r - Lesson | Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Math Courses ISTEP+ Grade 7 - Math: Test Prep & Practice Equations in the Form px + q = r and p(x + q) = r Contributors: Yuanxin (Amy) Yang Alcocer Instructor Instructor: Yuanxin (Amy) Yang Alcocer Show more Cite this lesson After reading this lesson, you'll learn how to solve algebraic problems such as px + q = r and p(x + q) = r easily. You'll also see what kinds of real world problems are represented by these equations. Table of Contents Solving for X in PX + Q = R Example of PX + Q = R Solving for X in P(X + Q) = R Examples of P(X + Q) = R Lesson Summary Show Create an account Video transcript VideoQuizCourse Click for sound 5:17 You must c C reate an account to continue watching Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Create Your Account To Continue Watching As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it now Already registered? Log in here for access Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. 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Student, Jefferson, Missouri Create an account Recommended lessons and courses for you Related LessonsRelated Courses ##### Multiplication Property of Equality | Overview, Example & Formula 4:05 ##### Addition Property of Equality | Definition, Explanation & Example 3:51 ##### Multi-Step Equations with Fractions & Decimals | Steps & Examples 6:41 ##### Division Property of Equality | Definition & Examples 3:51 ##### Subtraction Property of Equality | Overview & Examples 3:54 ##### How to Solve Equations on a Calculator 9:22 ##### Literal Equations | Definition, Formula & Examples 5:09 ##### Algebraic Model Definition, Applications & Examples 7:02 ##### Writing & Solving Multiplication Equations with One Variable 5:08 ##### Multiplication Principle | Definition, Equations & Examples 4:03 ##### Solving Equations Containing Parentheses 6:50 ##### Solving Multiple Step Equations | Explanation, Steps & Examples 5:44 ##### Solving Equations Using the Addition Principle 5:20 ##### Solving Equations Using Both Addition and Multiplication Principles 6:21 ##### Solving Division Word Problems with Two or More Variables 4:28 ##### Solving Addition Word Problems with Two or More Variables 8:57 ##### Solving Equations on the CLEP Scientific Calculator 5:30 ##### Algebraic Equation | Definition, Formula & Examples 6:16 ##### Solving Subtraction Word Problems with Two or More Variables 5:55 ##### Writing an Equation | Overview & Examples 5:22 ##### Smarter Balanced Assessments - Math Grade 6 Study Guide and Test Prep ##### Smarter Balanced Assessments - Math Grade 7 Study Guide and Test Prep ##### NES Middle Grades Mathematics (203) Study Guide and Test Prep ##### MEGA Middle School Mathematics Study Guide and Test Prep ##### ISEB Common Entrance Exam at 13+ Math: Study Guide & Test Prep ##### GED Math: Quantitative, Arithmetic & Algebraic Problem Solving ##### Study.com PSAT Study Guide and Test Prep ##### ORELA Middle Grades Mathematics Study Guide and Test Prep ##### WEST Middle Grades Mathematics (203) Study Guide and Test Prep ##### NMTA Middle Grades Mathematics (203): Practice & Study Guide ##### NMTA Essential Academic Skills Subtest Math (003): Practice & Study Guide ##### GACE Middle Grades Mathematics (013) Study Guide and Test Prep ##### OAE Middle Grades Mathematics (030) Study Guide and Test Prep ##### Smarter Balanced Assessments - Math Grade 8 Study Guide and Test Prep ##### MTTC Mathematics (Secondary) (022) Study Guide and Test Prep ##### Common Core Math Grade 7 - Expressions & Equations: Standards ##### Common Core Math Grade 6 - Expressions & Equations: Standards ##### NES Essential Academic Skills Subtest III Mathematics (003) Study Guide and Test Prep ##### CUNY Assessment Test in Math: Practice & Study Guide ##### Test for Admission into Catholic High Schools (TACHS) Study Guide and Exam Prep Solving for X in PX + Q = R --------------------------- Equations such as px + q = r are called linear equations because when graphed, they will give you a straight line. The x variable does not have an exponent with it, because linear equations never have exponents for their variables. You can easily solve equations of this form, along with equations of the form p(x + q) = r. In both equations, the p, q, and r represent rational numbers, either integers or fractions. It's really quite easy to solve the equation px + q = r. Remember, you want to isolate your x variable and move everything over to the other side. Here are your steps: Subtract q from both sides if positive, or add q to both sides if negative. This gives you px + q - q = r - q, which simplifies to px = r - q. Divide by p on both sides. This gives you px / p = (r - q) / p, which simplifies to x = (r - q) / p. Let's take a look at a real-world example of such a problem. Example of PX + Q = R --------------------- Say your local county fair has a base admission price of $5 per person. This is the price you pay just to get into the fair. Once you're in, you'll have to pay an additional $2 per ride. How many rides can you go on if you only have $15? Writing out the equation, you get 2 x + 5 = 15. q here represents your base admission, 5. p is your per ride price, 2. x is the number of rides you want to go on. r is the total admissions including the base price and rides, 15. Now, to find out how many rides you can get with $15, follow the steps. Subtract q from both sides. 2 x + 5 - 5 = 15 - 5 2 x = 10 Divide by p on both sides. 2 x / 2 = 10 / 2 x = 5 This tells you that you can get on 5 rides if you have $15. Solving for X in P(X + Q) = R ----------------------------- Now, let's see about solving problems that look like this: p(x + q) = r It's slightly different than the previous form. Notice the parentheses. This equation gives you the formula for the area of a rectangle when you know the length of one side and part of the length of the other side. The steps to solve for x are slightly different as well. Divide both sides by p. This gives you p(x + q) / p = r / p, which simplifies to x + q = r / p. Subtract q from both sides if positive, or add q to both sides if negative. This gives you x + q - q = r / p - q, which simplifies to x = r / p - q. Examples of P(X + Q) = R ------------------------ For example, say you want to build a shed that is 600 square feet around. The short sides of the shed will be 20 feet long, and the long sides will be 4 feet, plus some amount. How many more feet per side do you need to build to finish your shed? For this problem, your r is 600, your q is 4, and your p is 20. 20(x + 4) = 600 To solve it, follow your steps. First, divide by your p. In this problem, your p is 20. 20(x + 4) / 20 = 600 / 20 x + 4 = 30 Subtract q from both sides. In this problem, your q is 4. Since it's positive, we'll be subtracting from both sides. x + 4 - 4 = 30 - 4 x = 26 So you still need 26 feet for each of the sides that make up the length of the shed. Lesson Summary -------------- Let's review. Linear equations are equations that, when graphed, give you a straight line. Also, the variables in linear equations never have exponents. To solve the equation px + q = r, follow these steps: Subtract q from both sides (or add if negative). px + q - q = r - q px = r - q Divide by p on both sides. px / p = (r - q) / p x = (r - q) / p To solve the equation p(x + q) = r, follow these steps: Divide both sides by p. p(x + q) / p = r / p x + q = r / p Subtract q from both sides (or add if negative). x + q - q = r / p - q x = r / p - q Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. Teacher Try it now Go back Create an account to start this course today Start today. Try it now ISTEP+ Grade 7 - Math: Test Prep & Practice 22 chapters 171 lessons Chapter 1 ISTEP+ Grade 7 Math: Prime Factorization Prime Factorization | Definition, Methods & Examples 6:03 min Finding the Prime Factorization of a Number | Meaning & Examples 5:36 min Finding the Prime Factorization with Exponents 4:44 min Using Prime Factorizations to Find the Least Common Multiples 7:28 min Chapter 2 ISTEP+ Grade 7 Math: Square Roots Square Root | Definition, Formula & Examples 7:05 min Estimating Square Roots | Overview & Examples 5:10 min Simplifying Square Roots When not a Perfect Square 4:45 min Simplifying Square Root Expressions | Steps & Examples 7:03 min Perfect Square Definition, Formula & Special Properties 2:55 min Evaluating Square Roots of Perfect Squares 5:12 min Simplifying Square Roots | Overview & Examples 4:49 min Chapter 3 ISTEP+ Grade 7 Math: Rational & Irrational Numbers Rational Numbers | Definition, Forms & Examples 5:34 min Ordering & Comparing Rational Numbers | Steps, Tips & Examples 7:07 min Graphing Rational Numbers on a Number Line | Chart & Examples 5:02 min Irrational Numbers | Definition, Types & Examples 6:36 min Ordering & Graphing Irrational Numbers on a Number Line Chapter 4 ISTEP+ Grade 7 Math: Operations with Rational Numbers Adding & Subtracting Rational Numbers | Overview, Steps & Example 8:12 min Absolute Value | Definition & Examples 3:17 min Interpreting Sums of Rational Numbers Interpreting Differences of Rational Numbers Distributive Property in Math | Definition & Examples 6:20 min Arithmetic Calculations with Signed Numbers 5:21 min How Multiplication is Extended from Fractions to Rational Numbers Dividing Integers | Rules & Examples 6:03 min Chapter 5 ISTEP+ Grade 7 Math: Math Properties Identity Property: Concept & Examples | What is the Identity Property? 5:59 min Mathematical Properties | Definition & Examples 3:05 min Commutative Property | Definition, Examples & Applications 3:53 min Associative Property | Definition & Examples 4:28 min Additive Property of Inequality | Definition & Examples Multiplicative Identity | Definition, Explanation & Examples 3:36 min The Zero Property of Multiplication | Definition & Examples 2:40 min Distributive Property & Algebraic Expressions | Rules & Examples 5:04 min Chapter 6 ISTEP+ Grade 7 Math: Algebraic Expressions Variable in Algebra | Definition, Terms & Examples 5:26 min Algebraic Expression | Definition, Operations & Examples 5:12 min Evaluating Algebraic Expressions | Rules & Examples 7:27 min What is the Correct Setup to Solve Math Problems?: Writing Arithmetic Expressions 5:50 min Translating Words into Algebraic Expressions | Phrases & Examples 6:31 min Equivalent Expressions | Definition & Examples 4:42 min Combining Like Terms in Algebraic Expressions 7:04 min Solving Equations Using the Addition Principle 5:20 min Multiplication Principle | Definition, Equations & Examples 4:03 min Chapter 7 ISTEP+ Grade 7 Math: Graphing Basics Parts of a Graph | Labels & Examples 6:21 min Plotting Points on the Coordinate Plane 5:23 min Parabola | Definition, Formula & Examples 8:33 min Types of Parabolas | Overview, Graphs & Examples 6:15 min Chapter 8 ISTEP+ Grade 7 Math: Linear Equations Linear Equations | Definition, Formula & Solution 7:28 min Forms of a Linear Equation | Overview, Graphs & Conversion 6:38 min Abstract Algebraic Examples and Going from a Graph to a Rule 10:37 min Undefined & Zero Slope Graph | Definition & Examples 4:23 min Linear Equation | Parts, Writing & Examples 8:58 min Factoring in Algebra | Definition, Equations & Examples 5:32 min Equivalent Linear Expressions System of Equations in Algebra | Overview, Methods & Examples 8:39 min How Do I Use a System of Equations? 9:47 min How to Solve a System of Linear Equations in Two Variables 4:43 min How to Solve a Linear System in Three Variables With a Solution 5:01 min Solving System of Equations with 3 Variables | Steps & Examples 6:04 min Viewing now Equations in the Form px + q = r and p(x + q) = r 5:17 min Chapter 9 ISTEP+ Grade 7 Math: Inequalities Up next Inequality Signs in Math | Symbols, Examples & Variation 7:09 min Watch next lesson Inequality Notation | Overview & Examples 8:16 min Graphing Inequalities | Definition, Rules & Examples 7:59 min Graphing Inequalities | Overview & Examples 12:06 min Solve & Graph an Absolute Value Inequality | Formula & Examples 8:02 min Absolute Value Inequalities | Definition, Calculation & Examples 9:06 min Solving One-Step Inequalities | Definition, Steps & Examples 7:08 min Chapter 10 ISTEP+ Grade 7 Math: Slope & Rate of Change Slope | Definition, Formula & Examples 7:10 min Velocity and the Rate of Change 2:54 min Slopes and Rate of Change 3:11 min Determine the Rate of Change of a Function 5:40 min Constant Rate of Change | Definition, Formula & Examples 2:36 min Find the Slope of a Line | Formula & Examples 9:27 min Chapter 11 ISTEP+ Grade 7 Math: Ratios & Proportions Ratios & Rates | Differences & Examples 6:37 min Unit Rate in Math | Definition, Practice Problems & Solution 6:05 min Equivalent Ratios | Definition, Practice & Examples 4:33 min Proportion | Definition, Formula & Types 6:05 min Ratio & Proportion | Meaning, Differences & Examples 5:17 min Calculations with Ratios and Proportions 5:35 min How to Find an Unknown in a Proportion 6:06 min Proportional Relationship | Definition, Equation & Examples Graphing Proportional Relationships Proportional Relationships in Multistep Ratio & Percent Problems Chapter 12 ISTEP+ Grade 7 Math: Lines & Angles Line Segments & Rays | Differences & Measurement 3:59 min Parallel vs Perpendicular vs Transverse Lines Overview & Examples 6:06 min Parallel Lines and Transversals | Definition & Diagrams 7:40 min Acute Angle Definition, Types & Examples 1:45 min Obtuse Angle | Definition, Properties & Examples 2:35 min What is a Right Angle? - Definition & Formula 3:19 min Straight Angle | Definition, Characteristics & Examples 3:08 min Supplementary Angle | Definition, Properties, Theorem & Examples 4:29 min Vertical & Complementary Angles | Definition & Examples 4:17 min Chapter 13 ISTEP+ Grade 7 Math: Geometric Constructions Geometric Constructions Using Lines and Angles 4:32 min Line Segment Bisection & Midpoint Theorem: Geometric Construction 4:39 min Constructing a Parallel Line Using a Point Not on the Given Line 5:15 min Constructing Perpendicular Lines in Geometry 3:39 min Constructing an Angle Bisector in Geometry 3:36 min Constructing the Median of a Triangle 4:47 min Constructing Triangles: Types of Geometric Construction 5:59 min Chapter 14 ISTEP+ Grade 7 Math: Triangles Triangles in Geometry | Definition, Types & Formulas 4:30 min How to Classify Triangles | Overview & Examples 5:44 min Right Triangle | Properties, Proof & Theorems 5:58 min Interior & Exterior Angles of a Triangle | Overview & Examples 5:25 min Degrees in a Triangle | Measurement & Examples 5:14 min Altitude, Median & Angle Bisector of a Triangle 4:50 min ASA, SSS & SAS Triangle Postulates | Properties & Examples 6:15 min Angle-Angle-Side | Theorem, Proof & Example 6:31 min Chapter 15 ISTEP+ Grade 7 Math: Similar Figures Characteristics of a Polygon | Overview & Examples 4:25 min Similar Polygons Definition & Examples 8:00 min Constructing Similar Polygons 4:59 min Similar Triangles | Theorems, Formula & Examples 7:23 min Similar Triangles | Definition, Application Problems & Examples 6:23 min How to Read and Interpret Scale Drawings 4:27 min Scale Drawing Overview & Examples 5:42 min Chapter 16 ISTEP+ Grade 7 Math: Area & Circumference Pythagorean Theorem | Overview, Formula & Examples 7:33 min Area of Triangles & Rectangles | Formula, Calculation & Examples 5:43 min Area of a Parallelogram | Measurement, Formula & Examples 4:02 min Area of a Rhombus | Formula, Methods & Examples 6:30 min Area of a Trapezoid Formula & Examples 4:38 min Circle in Geometry | Definition, Parts & Examples 4:45 min Finding the Area & Circumference of a Circle 7:24 min Composite Figure | Overview & Examples 5:38 min Chapter 17 ISTEP+ Grade 7 Math: Complex Figures Planes and the Polyhedron: Definition and Example 3:52 min Platonic Solids | Definition, Properties & Types 4:39 min Three-Dimensional Shapes | Definition, Types & Characteristics 3:28 min Surface Area of a Cube & Rectangular Prism | Definition & Formula 4:08 min Surface Area of a Cylinder | Formula, Calculation & Examples 4:26 min Surface Area of a Pyramid | Formula, Calculation & Examples 5:11 min Volume of Prisms & Pyramid | Types, Formula & Calculation 6:15 min Volume of Cylinders, Cones & Spheres | Formula & Examples 7:50 min Constructing & Measuring Right Rectangular Prisms From Nets Chapter 18 ISTEP+ Grade 7 Math: Populations & Statistics Descriptive vs. Inferential Statistics | Definition & Examples 5:11 min Population vs Sample in Statistics | Differences & Examples 3:24 min Random Sampling Definition, Types & Examples 5:55 min Using Data from a Random Sample to Make Predictions Chapter 19 ISTEP+ Grade 7 Math: Data Analysis & Statistics What is the Center in a Data Set? - Definition & Options 5:08 min Spread of Data Overview & Examples 7:51 min Outliers in a Data Set | Minimums & Maximums 4:40 min Quartiles & Interquartile Range | Calculation & Examples 8:00 min Finding the Percentile of a Data Set | Formula & Example 8:25 min Standard Deviation Equation, Formula & Examples 13:05 min Data Collection Definition, Methods & Examples 5:20 min Overview on the Measures of Central Tendency in Psychology 6:00 min Measures of Central Tendency | Definition, Formula & Examples 8:30 min Skewed & Symmetric Distribution | Definition & Graphs 5:22 min Pie Chart vs. Bar Graph | Overview, Uses & Examples 9:36 min Creating & Interpreting Histograms: Process & Examples 5:43 min Box Plot | Definition, Uses & Examples 6:29 min Frequency Distribution | Definition, Graphs & Examples 5:15 min Chapter 20 ISTEP+ Grade 7 Math: Probability Probability of an Event | Simple, Compound & Complementary 6:55 min Conditional Probability | Overview, Calculation & Examples 5:10 min Independent & Dependent Events | Overview, Probability & Examples 12:06 min Probability of A or B | Overlapping & Non-Overlapping Events 7:05 min Probability of at Least One Event | Overview & Calculation 5:27 min Chance Probability | Overview, Examples & Calculation 5:21 min Probability & Relative Frequency Formula | Overview & Examples 5:53 min Chapter 21 ISTEP+ Grade 7 Math: Mathematical Process Critical Thinking and Logic in Mathematics 4:27 min Reasoning in Mathematics: Connective Reasoning 8:16 min Truth Value | Definition, Propositions & Tables 9:49 min Conjunction vs. Disjunction in Math | Overview & Characteristics 3:39 min Conditional Statement | Definition & Examples 4:54 min Conditional Statements | Converse, Inverse & Contrapositive 7:09 min Mathematical Problem | Definition, Principles & Uses 7:50 min The Three-Way Principle of Mathematics 5:49 min Using Mathematical Models to Solve Problems 6:35 min Chapter 22 ISTEP+ Grade 7 - Math Flashcards Related Study Materials Equations in the Form px + q = r and p(x + q) = r LessonsCoursesTopics ##### Multiplication Property of Equality | Overview, Example & Formula 4:05 ##### Addition Property of Equality | Definition, Explanation & Example 3:51 ##### Multi-Step Equations with Fractions & Decimals | Steps & Examples 6:41 ##### Division Property of Equality | Definition & Examples 3:51 ##### Subtraction Property of Equality | Overview & Examples 3:54 ##### How to Solve Equations on a Calculator 9:22 ##### Literal Equations | Definition, Formula & Examples 5:09 ##### Algebraic Model Definition, Applications & Examples 7:02 ##### Writing & Solving Multiplication Equations with One Variable 5:08 ##### Multiplication Principle | Definition, Equations & Examples 4:03 ##### Solving Equations Containing Parentheses 6:50 ##### Solving Multiple Step Equations | Explanation, 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7413
https://lessonplanet.com/teachers/khan-academy-multiply-decimals-3
Khan Academy: Multiply Decimals 3 Unit Plan for 4th - 6th Grade | Lesson Planet Search Search educational resources Search Menu Sign InTry It Free AI Teacher Tools Discover - [x] Discover Resources Search reviewed educational resources by keyword, subject, grade, type, and more Curriculum Manager (My Content) Manage saved and uploaded resources and folders To Access the Curriculum Manager Sign In or Join Now Browse Resource Directory Browse educational resources by subject and topic Curriculum Calendar Explore curriculum resources by date Lesson Planning Articles Timely and inspiring teaching ideas that you can apply in your classroom About - [x] Our Story Frequently Asked Questions Testimonials Contact Us Pricing School Access Your school or district can sign up for Lesson Planet — with no cost to teachers Learn More Sign In Try It Free Hi, What do you want to do? Create a lesson plan Generate resources with AI teacher tools Search 2 million educational videos Find a teaching resource Publisher Khan Academy Resource Details Curator Rating Educator Rating Not yet Rated Grade 4th - 6th SubjectsMath1 more... Resource TypeUnit Plans AudiencesFor Administrator Use2 more... Instructional StrategyIndependent Practice Lexile Measures0L Unit Plan Khan Academy: Multiply Decimals 3 Curated by ACT Multiply numbers with tenths and hundredths like 3.1x3.3 or 1.7x0.12 Students receive immediate feedback and have the opportunity to try questions repeatedly, watch a video or receive hints. 3 Views 0 Downloads CCSS:Adaptable Concepts multiplying decimals Show MoreShow Less Additional Tags decimal multiplication, khan academy, khan academy: multiply decimals 3, multiply decimals, multiply numbers with tenths and hundredths Show MoreShow Less Classroom Considerations Knovation Readability Score: 5 (1 low difficulty, 5 high difficulty) The intended use for this resource is Instructional|practice Common Core 5.NBT.B.7 See similar resources: Unknown Type #### Khan Academy: Multiplying Decimals 1 Khan Academy In this exercise, students practice multiplying decimals. Students receive immediate feedback and have the opportunity to get hints and try questions repeatedly. 3rd - 5th Math CCSS:Adaptable Unit Plan #### Khan Academy: Multiplying Decimals 3 (Standard Algorithm) Khan Academy Practice multiplying two numbers that are written to the tenths, hundredths, or thousandths place. Students receive immediate feedback and have the opportunity to try questions repeatedly, watch a video or receive hints. 5th - 7th Math Unknown Type #### Khan Academy: Multiplying Decimals 2 Khan Academy In this exercise, students practice multiplying decimals. Students receive immediate feedback and have the opportunity to get hints and try questions repeatedly. 3rd - 5th Math CCSS:Adaptable Unknown Type #### Khan Academy: Multiplying Decimals 3 (Standard Algorithm) Khan Academy In this exercise, students practice multiplying decimals 3. Students receive immediate feedback and have the opportunity to get hints and try questions repeatedly. 6th - 8th Math CCSS:Adaptable Unknown Type #### Khan Academy: Multiplying Decimals 1 (Standard Algorithm) Khan Academy Multiply a whole number times a decimal written to the tenths or hundredths place. Students receive immediate feedback and have the opportunity to try questions repeatedly, watch a video or receive hints. 4th - 6th Math CCSS:Adaptable Unit Plan #### Khan Academy: Multiplying Decimals 2 (Standard Algorithm) Khan Academy Multiply two numbers. Factors are written to the ones, tenths, or hundredths place. Students receive immediate feedback and have the opportunity to try questions repeatedly, watch a video or receive hints. 5th - 7th Math Unit Plan #### Khan Academy: Divide Decimals 3 Khan Academy Dividing decimals by whole numbers like 8.8/4. Students receive immediate feedback and have the opportunity to try questions repeatedly, watch a video or receive hints. 4th - 6th Math CCSS:Designed Instructional Video #### Khan Academy: Multiplying Decimals: Multiplication 8: Multiplying Decimals Khan Academy This video shows how to multiply decimals. 3rd - 8th Math CCSS:Adaptable Instructional Video #### Khan Academy: Multiplying Decimals: Multiplying a Decimal by a Power of 10 Khan Academy This video shows how to multiply a decimal by a power of 10. 3rd - 8th Math CCSS:Adaptable Unit Plan #### Khan Academy: Multiply by 3 Khan Academy Multiply 3 times a number less than or equal to 10. Students receive immediate feedback and have the opportunity to try questions repeatedly, watch a video or receive hints. CCSS.Math.Content.3.OA.C.7 Fluently multiply and divide within 100 2nd - 4th Math CCSS:Designed Try It Free © 1999-2025 Learning Explorer, Inc. 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7414
https://fermat.dartmouth.edu/posters/ugrad/2025/Marks-Alkire-poster.pdf
Fitting Regressions to Eigenvalues of Graphs Matthew Marks and Samuel Alkire Department of Mathematics, Dartmouth College Abstract Spectral graph theory connects graph structure to the eigenvalues of its adjacency and Laplacian matrices. In this project, we analyze how these eigenvalues evolve as the number of vertices increases across various graph families. We validate known formulas for path and cycle graphs, then extend our analysis to k-regular, complete, and k-partite graphs. Using regression techniques, we approximate spectral values and explore new families such as tailed graphs, where we observe that the maximum Laplacian eigenvalue grows linearly with the number of tails. Our findings offer tools for fast approximation and applications in network science and chemistry, while suggesting future directions for formal proofs and bounded-error estimates. Introduction and Literature Review Spectral Graph Theory Spectral graph theory is the study of the structural properties of graphs through the eigenvalues and eigenvectors of matrices associated with them, most notably the adjacency matrix and the Laplacian matrix. These spectral characteristics encode deep information about graph connectivity, expansion, and symmetry. In this project, we investigate how their spectral properties evolve as the number of vertices increases. Our primary goals are to: Our primary goals are to: • Uncover relationships between the eigenvalues of adjacency and Laplacian matrices across various graph families, • Analyze how these spectra evolve as the number of vertices increases, providing insight into the asymptotic behavior of large or growing graphs. Figure 1: Example of Different Types of Graphs Explored Literature Review We started our research by conducting a thorough literature review of the known relationships. Below are two graphs that we created illustrating the relationships 𝜆! 𝐿= 2 + 2 cos "! for the Adjacency Matrix of a Path Graph and 𝜆! 𝐴= 2 cos "! where n is the number of vertices and k is the index of the eigenvalue. Figure 2: Eigenvalues of Adjacency Figure 3: Eigenvalues of Laplacian Useful Formulas • Perron Frobenius Theorem for Symmetric Matrices • Let G be a connected undirected graph with eigenvalues μ1 ≥ μ2 ≥ ...μn -The eigenvalue μ1 has a strictly positive eigenvector -μ1 ≥ −μn -μ1 > μ2 • Max Eigenvalue Lemma • The maximum eigenvalue is bounded by the average degree and maximum degree of all vertices (dave ≤ λmax ≤ dmax) Conclusions and Next Steps Our regression-based approach reveals consistent patterns in how graph structure influences spectral properties. By confirming known results and extending them to new families like tailed and k-partite graphs, we demonstrate that simple structural features can predict eigenvalue behavior with surprising accuracy. Applications 1. Fast Approximation • Computing eigenvalues for large graphs can take O(n3) time, becoming extremely expensive for large graphs that model complex networks and datasets • Our regression would allow for a fast approximate solution, potentially in constant O(1) time, to this problem 2. Epidemic Modelling • In certain models, the chance of an epidemic dying out in a network is bounded by ) $ %!"#meaning that a larger max eigenvalue is correlated with a higher risk of outbreak • Thus we could model solutions to minimize the chances of a pandemic through removing nodes (vaccination), removing edges (quarantine communities), etc. Molecular Graphs for Chemistry • In chemistry, molecules can be modelled as graphs where vertices represent atoms and edges represent the chemical bonds between them • Eigenvalues correspond to molecular orbit, meaning the spectral gap between them can imply stability/instability Next Steps While our regression analysis provided us with estimations to the eigenvalues of these graphs, there is still a grey area on how effective our methods are. Firstly, we would hope to attempt to prove some of the relationships between the graphs and eigenvalues, such as seen in literature for cycle/path graphs. Finding an exact formula would be key to solving this problem. If unable to, we would then hope to understand how effective our analysis is in predicting these values by calculating an approximation factor to bound the effectiveness of our regression. This would be necessary so that anyone applying our approximate methods would be aware of possible errors in the calculation. Acknowledgements We would like to thank Professor Rosa Orellana for her invaluable guidance and mentorship throughout this project. We are also grateful to the James O. Freedman Presidential Scholars Program at Dartmouth College for providing the support and opportunity to pursue this research. References ref 1: Image from Wikipedia contributors. "ADE classification." Wikipedia, The Free Encyclopedia. ref 2: Lee, Shyi-Long & Yeh, Yeong-nan. (1993). On Eigenvalues and Eigenvectors of Graphs. Journal of Mathematical Chemistry. 12. 121-135. 10.1007/BF01164630. ref 3: Dadedzi, Kenneth. Analysis of Tree Spectra. PhD dissertation, University of Stellenbosch, 2018 ref 4: Chakrabarti, D., Wang, Y., Wang, C., Leskovec, J., & Faloutsos, C. (2008). Epidemic thresholds in real networks. ACM Transactions on Information and System Security (TISSEC), 10(4), Article 1. ref 5: Gutman, I., & Polansky, O. E. (1986). Mathematical Concepts in Organic Chemistry. Springer-Verlag. ref 6: Research Gate4 ref 7: Science Direct ref 8: Research Gate Our Work Max Eigenvalue of k-Regular Graphs A k-regular graph is a graph where all vertices have degree k, implying that the average degree equals the max degree. Using the lemma that the max eigenvalue is bounded by average/max degree, we found that for any k-regular graph λmax = k. This can be applied further to the following families: Cycle Graphs: All vertices have degree 2, implying a max eigenvalue of 2 for any size cycle. Figure 4: Cycle Graphs of size 3 to 6 Figure 5: Eigenvalues of Cycles Complete graphs: For a complete graph of size n, all vertices have degree n-1 implying a max eigenvalue of n-1 Figure 6: Complete Graphs of size 1 to 5 Figure 7: Max Eigenvalues of Complete graphs K-Partite Complete Graphs: If we have a complete k-partite graph where all k-components have size n, then all vertices have degree and max eigenvalue n(k-1) Figure 8: 5-Partite Complete Graph Figure 9: Max Eigenvalues of k-Partite Complete Graphs (k on x-axis, n on y-axis: results in a parabolic surface) Tailed Graphs To explore how local structural changes affect spectral properties, we examine path graphs with increasing numbers of tail vertices attached to one end. These "tail-extended" graphs preserve the linear backbone of the path but introduce degree-1 vertices that alter the boundary conditions. As more tails are added, we observe notable changes in the Eigenvalues of both the Adjacency and Laplacian Matrices. Figure 10: Eigenvalues of Laplacian Figure 11: Bounds for Laplacian Figure 11: Linear for increasing tails. as tails increase Approximations Figure 10 show the maximum eigenvalues of the Laplacian matrix as the number of vertices in the graph increases, color-coded by number of tails. We see that for each number of tails the maximum eigenvalue appears to approach an upper bound as the number of vertices increases. We plot this bound against number of tails in Figure 11 which demonstrates that the maximum eigenvalue seems to approach n+2 where n is the number of tails. Finally, in Figure 11 we estimate linear approximations for the maximum eigenvalue using number of tails as our independent variable for graphs with the same number of total vertices.
7415
https://www.youtube.com/watch?v=sWTiaaQ9xGk
Coordinate Geometry: Locus- Locus of P if PA + PB = 8 Jatheen's Math Channel 23800 subscribers 51 likes Description 4474 views Posted: 22 Mar 2017 Coordinate Geometry: Locus- Locus of P if PA + PB = 8 10 comments Transcript: Hello friends, welcome back. Hearty welcome to you. In this session, let's look at some more examples related to locus. As part of that, we will try to solve this problem. Find the equation of locus of point b. If a 2a 3 b 2 - 3 and given p a + pb = 8 we have to find the equation of locus of p given to us coordinates of a coordinates of b and we were told that the point for which we have to find out the locus its distance from a and its distance from b adds up to sums up to 8. By this definition itself we understand that it's an ellipse. We'll try to find out how the equation of locus will be. It will be an ellipse. We'll try to find out what it is. So first what we will do? We'll find out what is pa pb then we will sum it up and add up to 8. What is the distance formula which we'll be using? X2 - x1² plus y2 - y1 square under root square root will give us the distance between two points. So let's write down first what is PA? PA equal to let us say first the point is let P be a point on the locus with the coordinates X1 comma Y1. So PA will be square root of X1 - 2 square + Y1 - 3 square. Similarly, PB will be square root of X1 - 2² + Y1 minus of - 3. So, Y1 + 3². So, that is P and PB. Now, we have been given PA + PB equal to 8. Now when you try to add P and PB if you keep both of them on the left hand side it becomes complicated because you have to square it up it will be PA² + PB² + 2 PAB equal to 8. So what we will do? We'll try to take out PB to the other side and we'll write down PA = 8 minus PB. We'll square on both sides. So I'll have PA A² = 64. We are doing A minus B whole square formula plus PV² minus 16 into PV. Now we can substitute the values whatever we have P A² that means square root will go away. Can I write down x - x1 - 2² + y1 - 3² = 64 + pb² that is again x1 - 2² + y1 + 3² - 16 into PB. PB is square roo of x1 -1 to x1 - 2 square + y1 + 3 square. Now let's try to simplify this. When we go to the next step, we have x1 - 2 square x1 - 2 square cancel and we can see that this is of the form of a minus b whole square here and a + b whole square here. So I'll write down one step where we will say so that's nothing but y1 - 3² minus of y1 + 3² - 64 = -16 into so the point is very simple we just want to keep only the root here remaining whatever is left out we want to move to the left hand side so 16 into x1 - 2² + y1 + 3 square. Let's go to the next step. The next step will be we have a - b² minus of a + b whole square. So can I write it as - 4 a b - 4 into a is 3 into b = y1 - 4 a b. After that we have - 64 is equal to minus of 16 into square roo of x1 - 2² + y1 + 3 square. Now let's square it on both sides. Before that we can cancel. Let us see what I can cancel. I can cancel - I can cancel four times. four will go four times. Now let's square it on both sides. We'll have so I have to cancel four there. I should also cancel four from this one also. Or shall we take common and do it in the next step without confusing. Okay fine let's do that. So the next step what I will write down I will take -4 common. I will have 3 y1 + 16 = -4 into 4 into square roo of x1 - 2² + y1 + 3 all square. Now I can cancel -4 on both sides and we will square it up now. So we'll have next step 3 y1 + 16 all square is equal to 16 into the square root will go away now. So we'll have square 16 into x1 - t² + y1 + 3 square. Let's go to the next step. Next step will be a + b square. This is in the form of a + b whole square and left hand side we will square it up. A + b whole square formula we will apply not squaring we'll have 9 y1 squared + b² is 256 + 2 a b that is 16 into 348 into 2 96 y1 is equal to let's open this up. We will have 16 into x1² + y1 2 - 4 x1 + 6 y1 + 4 + 9 13. So we'll open the brackets here. We'll have 16 x1 squared + 16 y1 squar - 64 x1 + 96 y1 + 28 that is what we have on that side and in this side we have 9 y1 squar + 256 + 96 y1. Let us see can we cancel anything here? We can cancel 96 y1 [Music] and 96 y1. We can cancel 96 y1 on both sides. So let's put the next step take taking everything to right hand side. We'll have 16 x1 2 + 16 y 2 - 9 y1 2 that is 16 x² + 7 y1 2 - 64 x1 + 28 - 256 is - 48 = 0. So this is the equation of the locus. But if you want we can also write it so that we know that it's in log it's a ellipse. For that what we will do we'll use completing the squares method. 16 x1 2 - 64 x1 + 7 y1² - 48 =0. Now can I write it as 16 into x1² - 4 x1 + 7 y1 2 = 48. So completing the squares method if I do I'll write down 16 into x1 2 - 4x + 4 so that I can make it as x1 - 2². So now what I have to add on the other side 16 into 4 64. So that will become 112. Now I can divide this by 112 so that I can get the answer. So I'll write it as 16 into x1 - 2² + 7 y1² = 112 or can I write down x1 - 2² by 7 + y1² by 16 = 1. So now if we remove if we remove x1 y1 and make it as generic we'll have x - 2² by 7 + y² by 16 = 1. That is the equation of the locus of point p.
7416
https://www.youtube.com/watch?v=4ChbW830UQI
You can't assume the conclusion or how proof by contradiction works. Brunei Math Club 1570 subscribers Description 107 views Posted: 1 Sep 2022 By assuming the conclusion of a proposition to be proved, you can prove nothing. This video explains why it is so. Instead, we can assume the negation of the conclusion to prove by contradiction. See also: Transcript: all right in this video I will explain why we cannot assume the conclusion of a theorem of or preposition in a proof this seems quite trivial but I keep seeing students making the same mistake over the Nova first of all let's see an example okay so here's a preposition we want to prove uh let's say M and N are some integers actually it can be any integers if M squared is equal to 2 N squared then m is even this preposition appears when we want to prove that square root of 2 is irrational but anyway we are just interested in this part okay okay so let's give a proof that makes absolutely no sense okay so this is a wrong proof nonsense okay so suppose m is even okay here we are actually assuming the conclusion suppose m is even then that means m is equal to 2K for some uh K integer okay then let's substitute this into this equation then we have uh so 2K squared is 4K squared equal to 2 N squared so compare this so this is again wrong uh argument so compare this with this we have uh M squared is equal to 4K squared actually we we came back to our assumption here therefore m is equal to 2K actually this is wrong because if this is true then we should have m equal to plus minus 2K but anyway it doesn't matter I mean it is just simply everything is wrong so we're just adding another wrong point this indicates therefore m is even right and proof is done this makes no sense it's a nonsense but I keep finding this kind of proof uh in my class so what is exactly wrong with this proof okay so let's go back to our original proposition that is to be proved this one okay so this says if this is true then this is true okay so let's put uh this as a formal uh into a formal logical statement so let's say p is a proposition that M squared is equal to 2 N squared and Q is another proposition that m is even so what we want to prove is actually if P then Q so we want to prove this one however look at this wrong proof in the beginning we assume that m is even which is a conclusion and then we concluded that m is even right so what we are actually doing in this wrong proof is that if Q then Q so if you write down the truth table of this logical proposition we have this Q and Q implies Q so Q can be true or false okay when T Q is true then this is true and this is true therefore this implication is true you know this is according to the definition of this implication right and if Q is false then the premise here is false so so whatever and also the conclusion is false of course but if the premise is false then this whole proposition is true okay so this means this proposition if Q then Q or Q implies Q is always true regardless of the truth value of Q itself okay so such a logical statement is called tautology and this has no new information whatsoever and and it's quite useless why is it useless because you know we wanted to prove that m is even but this statement this tautology is true even if Q is false right so that means even if m is not even or that is if even if m is odd still this statement is true okay so this proof is in a sense true irrespective of the validity of this conclusion okay so this doesn't in other words this doesn't prove anything Okay so that's why this is wrong okay then what is the correct proof let's show that this one is correct okay first suppose suppose m is odd okay not even you know this is suppose this is odd we prove our contradiction actually then we have m is equal to 2 K plus 1 for some integer k okay then from the given equation we have 2K plus 1 squared is equal to 2 N squared and if we expand the left hand side we have 4 well K squared plus K plus 1 equal to 2 N squared now the right hand side is even obviously it's a multiple of two however the left hand side so this part is even an even number plus one is odd so the left hand side is odd but this is a contradiction so this says odd number is equal to an even number this cannot happen therefore our assumption was wrong that means this assumption was wrong therefore m is even so this is proof by contradiction so why is this proof correct let's go back to uh this statement so if we Define propositions q and P as this what we want to prove is this okay if P then Q however this uh proposition is equivalent to its contrapositive so not Q then not P okay so not Q here not Q in this case means not uh m is even so that means m is not even so that means so this means m is odd that is actually the Assumption we have made in the beginning of this proof and from that we derived that not p so not p in this case means so p is this one so naught p in this case means uh M squared is not equal to n 2 N squared so that is actually what we have derived here you know assuming m is odd then we derive that M squared becomes an odd number which cannot be equal to an even number so therefore M squared if it is if m is odd cannot be equal to 2N right so this this shows that M squared is not equal to 2 N Square it's it just cannot happen therefore we prove that this implication holds which means we have proved the original preposition so that's how proof by contradiction works
7417
https://algebra2.thinkport.org/module1/which-model-makes-sense-page-7.html
Module 1:Linear, Quadratic and Exponential Regression Skip to section navigationSkip to main content Module 1:Linear, Quadratic and Exponential Regression HomeStudent ResourcesTeacher Resources Home Which Model Makes Sense? Page 1 Page 2 Page 3 Page 4 Page 5 Page 6 Page 7 Page 8 Which Model Makes Sense? Home>Which Model Makes Sense?> Page 7 Resources for this lesson: > Glossary > Calculator Resources > Teacher Resources: Instructional Notes Compare your response with the solution below: Create and Analyze Upon initial inspection, a linear model appears to be a good fit. Test and Confirm The correlation coefficient is 0.981 and the scatter plot looks fairly linear. However, when examining the residual plot, there is a clear pattern to the residuals, meaning that a linear model is not the best fit. While the quadratic curve seems to fit the data well, the residual plot reveals a clear pattern. This indicates that a quadratic model is not the best fit. Since the data is continuing to increase, the only other model that makes sense is an exponential model. The exponential curve of best fit equation is y = 35.156(1.023)x. The residuals for the exponential model are: | Years since 1950 | 0 | 10 | 20 | 30 | 40 | 50 | 60 | | Residual value (in thousands) | 0.080 | 0.093 | 0.106 | −0.469 | −0.172 | 0.369 | −0.011 | The residual graph is random and close to zero, confirming that the exponential model is a good fit for the data. It appears that an exponential model is the best fit for this data set. < PreviousNext > Site Map|Accessibility|About This website is a production of Maryland Public Television/Thinkport in collaboration with the Maryland State Department of Education. The contents of this website were developed under a grant from the U.S. Department of Education. However, those contents do not necessarily represent the policy of the U.S. Department of Education, and you should not assume endorsement by the Federal Government. 2014 Copyright Maryland State Department of Education Contact the MSDE Office of Instructional Technology for copyright questions.
7418
https://us.sofatutor.com/math/videos/equivalent-decimal-fractions
To make full use of our website, enable JavaScript in your browser. Our tutorial videos allow students to learn at their own pace without any pressure or stress. They can watch, pause or rewind the videos as often as they need until they understand the content. Our interactive exercises come in a variety of formats so students can practice in a playful way. They get feedback and hints while solving the exercises. As a result, they learn and grow from their mistakes instead of feeling frustrated. Children in elementary school can use Sofahero to review independently and stay motivated. They master topics while going on exciting adventures, without the help of adults. Students can use the worksheets to prepare themselves for mock tests: Simply print them out, fill them out, and check the answers using the answer key. With interactive e-books children can playfully train their active listening and comprehension skills. Discover why over 1.6 MILLION students choose sofatutor! Equivalent Decimal Fractions Watch videos Start exercises Show worksheets Unlock this video in just a few steps, and benefit from all sofatutor content: You must be logged in to be able to give a rating. Wow, thank you! Please rate us on Google too! We look forward to it! Basics on the topic Equivalent Decimal Fractions Equivalent Decimal Fractions A fraction is a portion or part of any quantity out of a whole. There are different types of fractions including: improper fractions, mixed fractions, proper fractions, and decimal fractions. In this video we will learn about decimal fractions and how to identify equivalent decimal fractions as well as how to write decimal fractions. We will show a couple of examples after which you can practice a set of decimal fractions problems with ease. Revision – Decimal Fractions Let’s recap what decimal fractions are first: Decimal fractions are fractions where the denominator is a power of ten. A power of ten means multiplying ten by itself a certain number of times. For example, 10x10 = 100, and 10x10x10 =1000. Equivalent Decimal Fractions – Definition Equivalent decimal fractions are fractions that are equal in value and the denominator is a power of ten. For example, fractions with denominators of 10, 100, 1000 and so on. Comparing Decimal Fractions – Examples A great visual method to compare decimal fractions is to use the base 10 blocks. Below are some examples. Comparing Decimal Fractions – Example 1 In this first base 10 block the whole is divided into ten equal parts, so the denominator is ten. Four out of ten parts are shaded in blue; so we label the shaded parts in the numerator as four, making the decimal fraction $\frac{4}{10}$. On the other base 10 block, the whole is divided into one hundred equal parts, so the denominator is one hundred. Forty sections out of one hundred parts are shaded in red which we can represent as the decimal fraction $\frac{40}{100}$. The base 10 blocks below help us when comparing fractions with denominators of 10 and 100. We can convert fractions with denominators of 10 and 100 to decimals. Our $\frac{4}{10}$ and $\frac{40}{100}$ can be written as 0.4 = 0.40. | | Denominator 10 | Denominator 100 | --- | fraction | $\frac{4}{10}$ | $\frac{40}{100}$ | | decimal | 0.4 | 0.40 | Comparing Decimal Fractions – Example 2 Let’s look at different examples to clarify our understanding of equivalent fractions with denominators of 10 and 100. Equivalent fractions and decimals with denominators of 10 and 100 are more comprehensible if we visualize them. In this base 10 block the whole is divided into one hundred equal parts, and ninety out of one hundred parts are shaded in red which we can write as a decimal fraction $\frac{90}{100}$. In the second base 10 block the whole is divided into ten equal parts, and nine out of ten parts are shaded in blue, so we can represent the shaded parts as a decimal fraction $\frac{9}{10}$. The base 10 blocks represent two equivalent fractions with denominators of 10 and 100. | | Denominator 10 | Denominator 100 | --- | fraction | $\frac{9}{10}$ | $\frac{90}{100}$ | | decimal | 0.9 | 0.90 | Equivalent Decimal Fractions – Summary How do we compare decimal fractions? Look at the summary below: You can also look at our worksheets and videos on equivalent decimal fractions with denominators of 10 and 100 for more practice. We also provide worksheets on adding fractions with denominators of 10 and 100 as well as place value and decimal fractions worksheets. Soon you’ll be able to answer questions like “what is a fraction with a denominator of 10 ?” and “what is a fraction that has a denominator of 100?”easily! Frequently Asked Questions on Equivalent Decimal Fractions What are decimal fractions? Decimal fractions are fractions where the denominator is a power of ten: 10,100, etc. What are equivalent decimal fractions? Equivalent decimal fractions are fractions that are equal in value but still the denominator is a power of ten. Decimal fractions are fractions where the denominator is a power of ten: 10, 10, 100, etc. What are equivalent decimal fractions? Equivalent decimal fractions are fractions that are an equal in value but still the denominator is a power of ten. Transcript Equivalent Decimal Fractions "Doo dee do do doo (...) OH, what does this button do Axel?!" "Tank don't push that!" Axel and Tank need to get back home. However, they need to crack the codes to unlock the doors. Let's help them by identifying, Equivalent Decimal Fractions. A decimal fraction is a fraction with a denominator that is a power of ten. A power of ten is a number that can be formed by multiplying ten times itself... such as ten (...) or one hundred. Equivalent fractions are fractions that are EQUAL in value. We can compare decimal fractions using base ten blocks. For example, near the first door we see four tenths. This base ten block represents TENTHS, because the whole is broken up into ten equal parts... and since four out of ten parts are shaded the numerator is four. To unlock the door, we need to identify a decimal fraction that is equivalent to four tenths. The denominator of the equivalent fraction is one hundred, because we see HERE the whole is broken up into one hundred equal parts. That means this base ten block represents HUNDREDTHS. In order to identify the numerator, we shade the same VALUE, or amount, as four tenths on the base ten block. How many HUNDREDTHS are shaded in? (...) FORTY, because there are four columns with ten in each shaded. Four tenths is equal to forty hundredths because the same VALUE is shaded in... so Axel and Tank go to the next door. Here, we need to identify a decimal fraction that is equivalent to ninety hundredths. What is the denominator? (...) This whole is divided into tenths, so the denominator is ten. How many tenths do we shade in to make an equivalent fraction? (...) In order to be equivalent, the same VALUE must be shaded in, so we shade in NINE... (...) and write it HERE. Ninety hundredths is equal to nine tenths because the same VALUE is shaded in. Axel and Tank head to the last door. This time, we need to write the fraction represented on the base ten block and THEN find an equivalent decimal fraction. Try solving on your own. Pause the video so you have time to work (...) and press play when you're ready to see the steps for finding the answer! First, write seven tenths HERE because there are seven tenths shaded on the base ten block. Since there are one hundred parts in our whole, write the denominator one hundred HERE. Next shade in seventy hundredths because both blocks need to have the same VALUE in order to be equivalent. Seven tenths is equal to seventy hundredths. Before we see if Axel and Tank have made it back home, let's summarize. Remember (...) we can compare decimal fractions using base ten blocks. In order to identify equivalent decimal fractions, first write the denominator. Second, shade in the same value on the other base ten block. Last, count how much you shaded in, and write the result as the numerator. "Wait this isn't our home (...) but it looks a lot like it!" "Yeah really nice choice on the decor Mrs. Loch Ness Monster, (...) OH what does this button do?!?" "Tank, PLEASE not again!" Equivalent Decimal Fractions exercise What does equivalent mean when we are talking about fractions? Remember to look at the shading in the picture. Are they the same? Does one have more than the other? The fractions we see in the previous hint are equivalent to each other: $\dfrac{4}{10}$ is equivalent to $\dfrac{40}{100}$. Equivalent means equal to or the same as. When two fractions are different, but have the same value they are equivalent. Can you identify the correct base ten block? Remember, you are finding the fraction represented in tens or hundreds blocks. The number in the denominator determines how many boxes are in the picture. The numerator is represented by the amount of blocks shaded in. What is the fraction represented on the base ten block? Remember that the numerator represents the number of boxes shaded. The base ten block below illustrates $\frac{4}{10}$. How can this help you complete the above fractions? 1.The first image represents $\frac{8}{10}$. 8 of the 10 boxes are shaded in. 2.The second image represents $\frac{50}{100}$. 50 of the 100 boxes are shaded in. 3.The third image represent $\frac{80}{100}$. 80 of the 100 boxes are shaded in. 4.The fourth image represents $\frac{5}{10}$. 5 of the 10 boxes are shaded in. Match the equivalent fractions. These images will help you find one of the pairs above. They are equivalent fractions. Remember that the shaded part represents the numerator and the total number of boxes the denominator. Remember that we can also think of these numbers as being multiplied by 10. For example, $\frac{4}{10}$ = $\frac{40}{100}$ When we multiply 4 by 10 we get 40 and when we multiply 10 by 10 we get 100. These are equivalent fractions. $\frac{9}{10}$ = $\frac{90}{100}$ We can see on the shaded blocks that 9 of the 10 blocks are shaded, which looks the same as 90 of the 100 blocks that are also shaded. If we were to lay one on top of the other the same amount would be shaded in, making them equivalent. $\frac{6}{10}$ = $\frac{60}{100}$ $\frac{2}{10}$ = $\frac{20}{100}$ $\frac{1}{10}$ = $\frac{10}{100}$ What is the missing fraction? We can tell from looking at the base ten block images if this is an equivalent fraction or not. Remember the total number of boxes the square is divided into represents the denominator, or the number that goes in the bottom half of the fraction. The numerator is represented by the number of blocks shaded in. The second fraction represents our original fraction changed into a hundredth denominator. 30 out of 100 blocks are filled in, so $\frac{3}{10}$ becomes $\frac{30}{100}$. The two fractions are equivalent. How much pizza have Axel and Tank eaten after their adventure? Remember that the pizza is similar to the base 10 block. Look at the pizza represented in a base 10 block. Remember the numerator is the number of pieces eaten and the denominator is the total number of pieces of pizza. Count all of the pieces of the pizza. There are 10 in total. Of those 10, how many are missing? There are 4 pieces of pizza missing. That means that Axel and Tank ate 4 pieces out of 10 or, they ate $\frac{4}{10}$ of the pizza. $\dfrac{4}{10}$ = $\dfrac{40}{100}$ =. $\dfrac{400}{1000}$ Converting Decimals and Fractions Equivalent Decimal Fractions
7419
https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)/07%3A_Electric_Potential/7.05%3A_Determining_Field_from_Potential
Skip to main content 7.5: Determining Field from Potential Last updated : Mar 2, 2025 Save as PDF 7.4: Calculations of Electric Potential 7.6: Equipotential Surfaces and Conductors Page ID : 4389 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives By the end of this section, you will be able to: Explain how to calculate the electric field in a system from the given potential Calculate the electric field in a given direction from a given potential Calculate the electric field throughout space from a given potential Recall that we were able, in certain systems, to calculate the potential by integrating over the electric field . As you may already suspect, this means that we may calculate the electric field by taking derivatives of the potential, although going from a scalar to a vector quantity introduces some interesting wrinkles. We frequently need to calculate the force in a system; since it is often simpler to calculate the potential directly, there are systems in which it is useful to calculate and then derive from it. In general, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because the force on a positive charge is in the direction of and also in the direction of lower potential . Furthermore, the magnitude of equals the rate of decrease of with distance. The faster decreases over distance, the greater the electric field . This gives us the following result. Relationship between Voltage and Uniform Electric Field In equation form, the relationship between voltage and uniform electric field is where is the distance over which the change in potential takes place. The minus sign tells us that points in the direction of decreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electric potential . For continually changing potentials, and become infinitesimals, and we need differential calculus to determine the electric field . As shown in Figure , if we treat the distance as very small so that the electric field is essentially constant over it, we find that Therefore, the electric field components in the Cartesian directions are given by This allows us to define the “grad” or “del” vector operator, which allows us to compute the gradient in one step. In Cartesian coordinates, it takes the form With this notation, we can calculate the electric field from the potential with a process we call calculating the gradient of the potential. If we have a system with either cylindrical or spherical symmetry , we only need to use the del operator in the appropriate coordinates: Example : Electric Field of a Point Charge Calculate the electric field of a point charge from the potential. Strategy The potential is known to be , which has a spherical symmetry . Therefore, we use the spherical del operator (Equation ) into Equation : Solution Performing this calculation gives us This equation simplifies to as expected. Significance We not only obtained the equation for the electric field of a point particle that we’ve seen before, we also have a demonstration that points in the direction of decreasing potential, as shown in Figure . Example : Electric Field of a Ring of Charge Use the potential found previously to calculate the electric field along the axis of a ring of charge (Figure ). Strategy In this case, we are only interested in one dimension, the -axis. Therefore, we use with the potential found previously. Solution Taking the derivative of the potential yields Significance Again, this matches the equation for the electric field found previously. It also demonstrates a system in which using the full del operator is not necessary. Exercise Which coordinate system would you use to calculate the electric field of a ? Answer : Any, but cylindrical is closest to the symmetry of a dipole . 7.4: Calculations of Electric Potential 7.6: Equipotential Surfaces and Conductors
7420
https://epubs.siam.org/doi/10.1137/S0895479800368688
Evaluating Padé Approximants of the Matrix Logarithm | SIAM Journal on Matrix Analysis and Applications Skip to main content Search Search This Journal This Journal Anywhere Books Journals Proceedings Quick Search in Journals Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Quick Search anywhere Enter Search Terms Search Advanced Search 0 Register / Sign In Access via your Institution Skip main navigationClose Drawer Menu Open Drawer Menu Menu Journals SIAM Review Multiscale Modeling & Simulation SIAM Journal on Applied Algebra and Geometry SIAM Journal on Applied Dynamical Systems SIAM Journal on Applied Mathematics SIAM Journal on Computing SIAM Journal on Control and Optimization SIAM Journal on Discrete Mathematics SIAM Journal on Financial Mathematics SIAM Journal on Imaging Sciences SIAM Journal on Life Sciences SIAM Journal on Mathematical Analysis SIAM Journal on Mathematics of Data Science SIAM Journal on Matrix Analysis and Applications SIAM Journal on Numerical Analysis SIAM Journal on Optimization SIAM Journal on Scientific Computing SIAM/ASA Journal on Uncertainty Quantification Theory of Probability & Its Applications Locus E-books Bookstore Proceedings For Authors Journal Authors Book Authors ICM Authors For Librarians Collections Epidemiology Collection High Impact Article Collection JOIN SIAM HELP/CONTACT US Journal Home Current Issue All Issues About About this Journal Editorial Policy Editorial Board Instructions for Authors Instructions for Referees Submit Subscribe Share Share on Facebook X LinkedIn Email HomeSIAM Journal on Matrix Analysis and ApplicationsVol. 22, Iss. 4 (2001)10.1137/S0895479800368688 Previous articleNext article Evaluating Padé Approximants of the Matrix Logarithm Author: Nicholas J.HighamAuthors Info & Affiliations Get Access BibTeX Tools Add to favorites Download Citations Track Citations Contents ###### PREVIOUS ARTICLE Approximating the Logarithm of a Matrix to Specified Accuracy Previous###### NEXT ARTICLE Joint Approximate Diagonalization of Positive Definite Hermitian Matrices Next Abstract References Information & Authors Metrics & Citations Get Access References Figures Tables Media Share Abstract The inverse scaling and squaring method for evaluating the logarithm of a matrix takes repeated square roots to bring the matrix close to the identity, computes a Padé approximant, and then scales back. We analyze several methods for evaluating the Padé approximant, including Horner's method (used in some existing codes), suitably customized versions of the Paterson--Stockmeyer method and Van Loan's variant, and methods based on continued fraction and partial fraction expansions. The computational cost, storage, and numerical accuracy of the methods are compared. We find the partial fraction method to be the best method overall and illustrate the benefits it brings to a transformation-free form of the inverse scaling and squaring method recently proposed by Cheng, Higham, Kenney, and Laub [SIAM J. Matrix Anal. Appl., 22 (2001), pp. 1112--1125]. We comment briefly on how the analysis carries over to the matrix exponential. MSC codes 65F30 Keywords matrix logarithm Padé approximation inverse scaling and squaring method Horner's method \PS\ method continued fraction partial fraction expansion Get full access to this article View all available purchase options and get full access to this article. Get Access Sign in as an individual or via your institution References 1. G. A. Baker, Jr. and P. Graves‐Morris, Padé Approximants, 2nd ed., Encyclopedia Math. Appl., Cambridge University Press, Cambridge, UK, 1996. Google Scholar 2. ˙𝐴 ke Björck, Sven Hammarling, A Schur method for the square root of a matrix, Linear Algebra Appl., 52/53 (1983), 127–140 Crossref Web of Science Google Scholar 3. G. Blanch, Numerical evaluation of continued fractions, SIAM Rev., 6 (1964), 383–421 Abstract Web of Science Google Scholar 4. D. Calvetti, E. Gallopoulos, L. Reichel, Incomplete partial fractions for parallel evaluation of rational matrix functions, J. Comput. Appl. Math., 59 (1995), 349–380 Crossref Web of Science Google Scholar 5. S. H. Cheng, N. J. Higham, C. S. Kenney, and A. J. Laub, Approximating the logarithm of a matrix to specified accuracy, SIAM J. Matrix Anal. Appl., 22 (2001), pp. 1112–1125. Abstract Web of Science Google Scholar 6. P. J. Davis and P. Rabinowitz, Methods of Numerical Integration, 2nd ed., Academic Press, Orlando, FL, 1984. Google Scholar 7. Luca Dieci, Benedetta Morini, Alessandra Papini, Computational techniques for real logarithms of matrices, SIAM J. Matrix Anal. Appl., 17 (1996), 570–593 Abstract Web of Science Google Scholar 8. W. Gautschi, Algorithm 726: ORTHPOL—A package of routines for generating orthogonal polynomials and Gauss‐type quadrature rules, ACM Trans. Math. Software, 20 (1994), pp. 21–62. Crossref Web of Science Google Scholar 9. G. H. Golub and C. F. Van Loan. Matrix Computations, 3rd ed., Johns Hopkins University Press, Baltimore, MD, 1996. Google Scholar 10. Nicholas Higham, Computing real square roots of a real matrix, Linear Algebra Appl., 88/89 (1987), 405–430 Crossref Web of Science Google Scholar 11. Nicholas Higham, Accuracy and stability of numerical algorithms, Society for Industrial and Applied Mathematics (SIAM), 1996xxviii+688 Google Scholar 12. R. A. Horn and C. R. Johnson, Topics in Matrix Analysis, Cambridge University Press, London, 1991. Google Scholar 13. Charles Kenney, Alan Laub, Condition estimates for matrix functions, SIAM J. Matrix Anal. Appl., 10 (1989), 191–209 Abstract Web of Science Google Scholar 14. Charles Kenney, Alan Laub, Padé error estimates for the logarithm of a matrix, Internat. J. Control, 50 (1989), 707–730 Crossref Web of Science Google Scholar 15. Cleve Moler, Charles Van Loan, Nineteen dubious ways to compute the exponential of a matrix, SIAM Rev., 20 (1978), 801–836 Abstract Web of Science Google Scholar 16. Michael Paterson, Larry Stockmeyer, On the number of nonscalar multiplications necessary to evaluate polynomials, SIAM J. Comput., 2 (1973), 60–66 Abstract Google Scholar 17. W. H. Press, S. A. Teukolsky, W. T. Vetterling, and B. P. Flannery, Numerical Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed., Cambridge University Press, London, 1992. Google Scholar 18. Charles Van Loan, A note on the evaluation of matrix polynomials, IEEE Trans. Automat. Control, 24 (1979), 320–321 Crossref Web of Science Google Scholar 19. R. C. Ward, Numerical computation of the matrix exponential with accuracy estimate, SIAM J. Numer. Anal., 14 (1977), pp. 600–610. Abstract Web of Science Google Scholar Show all references Information & Authors Information Authors Information Published In SIAM Journal on Matrix Analysis and Applications Volume 22 • Issue 4 • January 2001 Pages: 1126 - 1135 DOI: 10.1137/S0895479800368688 ISSN (online): 1095-7162 Copyright Copyright © 2001 Society for Industrial and Applied Mathematics. History Published online: 31 July 2006 MSC codes 65F30 Keywords matrix logarithm Padé approximation inverse scaling and squaring method Horner's method \PS\ method continued fraction partial fraction expansion Authors Affiliations Expand All Nicholas J.Higham View all articles by this author Metrics & Citations Metrics Citations Metrics Metrics Downloads Citations No data available. 269 34 Total 6 Months 12 Months Total number of downloads Citations If you have the appropriate software installed, you can download article citation data to the citation manager of your choice. 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Relton SIAM Journal on Scientific Computing, Vol. 35, No. 4 | 24 July 2013 AbstractPDF (495 KB) Improved Inverse Scaling and Squaring Algorithms for the Matrix Logarithm Awad H. Al-Mohyand Nicholas J. Higham SIAM Journal on Scientific Computing, Vol. 34, No. 4 | 24 July 2012 AbstractPDF (278 KB) A Schur–Padé Algorithm for Fractional Powers of a Matrix Nicholas J. 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Daviesand Nicholas J. Higham SIAM Journal on Matrix Analysis and Applications, Vol. 25, No. 2 | 31 July 2006 AbstractPDF (240 KB) Approximating the Logarithm of a Matrix to Specified Accuracy Sheung Hun Cheng, Nicholas J. Higham, Charles S. Kenney,and Alan J. Laub SIAM Journal on Matrix Analysis and Applications, Vol. 22, No. 4 | 31 July 2006 AbstractPDF (176 KB) Figures Tables Media Share Share Copy the content Link Copy Link Copied! Copying failed. Share with email Email a colleague Share on social media FacebookX (formerly Twitter)LinkedInemail Get Access Get Access PurchaseSave for laterItem saved, go to cart Article Pay-Per-View $40.00 Add to cart Article Pay-Per-View Checkout Access via your Institution Questions about how to access this content? Contact SIAM at service@siam.org. References References 1. G. A. Baker, Jr. and P. Graves‐Morris, Padé Approximants, 2nd ed., Encyclopedia Math. Appl., Cambridge University Press, Cambridge, UK, 1996. Google Scholar 2. ˙𝐴 ke Björck, Sven Hammarling, A Schur method for the square root of a matrix, Linear Algebra Appl., 52/53 (1983), 127–140 Crossref Web of Science Google Scholar 3. G. Blanch, Numerical evaluation of continued fractions, SIAM Rev., 6 (1964), 383–421 Abstract Web of Science Google Scholar 4. D. Calvetti, E. Gallopoulos, L. Reichel, Incomplete partial fractions for parallel evaluation of rational matrix functions, J. Comput. Appl. Math., 59 (1995), 349–380 Crossref Web of Science Google Scholar 5. S. H. Cheng, N. J. Higham, C. S. Kenney, and A. J. Laub, Approximating the logarithm of a matrix to specified accuracy, SIAM J. Matrix Anal. Appl., 22 (2001), pp. 1112–1125. Abstract Web of Science Google Scholar 6. P. J. Davis and P. Rabinowitz, Methods of Numerical Integration, 2nd ed., Academic Press, Orlando, FL, 1984. Google Scholar 7. Luca Dieci, Benedetta Morini, Alessandra Papini, Computational techniques for real logarithms of matrices, SIAM J. Matrix Anal. Appl., 17 (1996), 570–593 Abstract Web of Science Google Scholar 8. W. Gautschi, Algorithm 726: ORTHPOL—A package of routines for generating orthogonal polynomials and Gauss‐type quadrature rules, ACM Trans. Math. Software, 20 (1994), pp. 21–62. Crossref Web of Science Google Scholar 9. G. H. Golub and C. F. Van Loan. Matrix Computations, 3rd ed., Johns Hopkins University Press, Baltimore, MD, 1996. Google Scholar 10. Nicholas Higham, Computing real square roots of a real matrix, Linear Algebra Appl., 88/89 (1987), 405–430 Crossref Web of Science Google Scholar 11. Nicholas Higham, Accuracy and stability of numerical algorithms, Society for Industrial and Applied Mathematics (SIAM), 1996xxviii+688 Google Scholar 12. R. A. Horn and C. R. Johnson, Topics in Matrix Analysis, Cambridge University Press, London, 1991. Google Scholar 13. Charles Kenney, Alan Laub, Condition estimates for matrix functions, SIAM J. Matrix Anal. Appl., 10 (1989), 191–209 Abstract Web of Science Google Scholar 14. Charles Kenney, Alan Laub, Padé error estimates for the logarithm of a matrix, Internat. J. Control, 50 (1989), 707–730 Crossref Web of Science Google Scholar 15. Cleve Moler, Charles Van Loan, Nineteen dubious ways to compute the exponential of a matrix, SIAM Rev., 20 (1978), 801–836 Abstract Web of Science Google Scholar 16. Michael Paterson, Larry Stockmeyer, On the number of nonscalar multiplications necessary to evaluate polynomials, SIAM J. Comput., 2 (1973), 60–66 Abstract Google Scholar 17. W. H. Press, S. A. Teukolsky, W. T. Vetterling, and B. P. Flannery, Numerical Recipes in FORTRAN: The Art of Scientific Computing, 2nd ed., Cambridge University Press, London, 1992. Google Scholar 18. Charles Van Loan, A note on the evaluation of matrix polynomials, IEEE Trans. Automat. Control, 24 (1979), 320–321 Crossref Web of Science Google Scholar 19. R. C. Ward, Numerical computation of the matrix exponential with accuracy estimate, SIAM J. Numer. Anal., 14 (1977), pp. 600–610. Abstract Web of Science Google Scholar Recommended Content Approximating the Logarithm of a Matrix to Specified Accuracy Sheung Hun Cheng , Nicholas J. Higham , Charles S. Kenney , Alan J. Laub Abstract The standard inverse scaling and squaring algorithm for computing the matrix logarithm begins by transforming the matrix to Schur triangular form in order to facilitate subsequent matrix square root and Padé approximation computations. A transformation-free form of this method that exploits incomplete Denman--Beavers square root iterations and aims for a specified accuracy (ignoring roundoff) is presented. The error introduced by using approximate square roots is accounted for by a novel splitting lemma for logarithms of matrix products. The number of square root stages and the degree of the final Padé approximation are chosen to minimize the computational work. This new method is attractive for high-performance computation since it uses only the basic building blocks of matrix multiplication, LU factorization and matrix inversion. The Scaling and Squaring Method for the Matrix Exponential Revisited Nicholas J. Higham Abstract The scaling and squaring method is the most widely used method for computing the matrix exponential, not least because it is the method implemented in MATLAB's {\tt expm} function. The method scales the matrix by a power of 2 to reduce the norm to order 1, computes a Padé approximant to the matrix exponential, and then repeatedly squares to undo the effect of the scaling. We give a new backward error analysis of the method (in exact arithmetic) that employs sharp bounds for the truncation errors and leads to an implementation of essentially optimal efficiency. We also give new rounding error analysis that shows the computed Padé approximant of the scaled matrix to be highly accurate. For IEEE double precision arithmetic the best choice of degree of Padé approximant turns out to be 13, rather than the 6 or 8 used by previous authors. Our implementation of the scaling and squaring method always requires at least two fewer matrix multiplications than {\tt expm} when the matrix norm exceeds 1, which can amount to a 37% saving in the number of multiplications, and it is typically more accurate, owing to the fewer required squarings. We also investigate a different scaling and squaring algorithm proposed by Najfeld and Havel that employs a Padé approximation to the function $x \coth(x)$. This method is found to be essentially a variation of the standard one with weaker supporting error analysis. Improved Inverse Scaling and Squaring Algorithms for the Matrix Logarithm Awad H. Al-Mohy , Nicholas J. Higham Abstract A popular method for computing the matrix logarithm is the inverse scaling and squaring method, which essentially carries out the steps of the scaling and squaring method for the matrix exponential in reverse order. Here we make several improvements to the method, putting its development on a par with our recent version [SIAM J. Matrix Anal. Appl., 31 (2009), pp. 970--989] of the scaling and squaring method for the exponential. In particular, we introduce backward error analysis to replace the previous forward error analysis; obtain backward error bounds in terms of the quantities $\|A^p\|^{1/p}$, for several small integer $p$, instead of $\|A\|$; and use special techniques to compute the argument of the Padé approximant more accurately. We derive one algorithm that employs a Schur decomposition, and thereby works with triangular matrices, and another that requires only matrix multiplications and the solution of multiple right-hand side linear systems. Numerical experiments show the new algorithms to be generally faster and more accurate than their existing counterparts and suggest that the Schur-based method is the method of choice for computing the matrix logarithm. A Schur–Padé Algorithm for Fractional Powers of a Matrix Nicholas J. Higham , Lijing Lin Abstract A new algorithm is developed for computing arbitrary real powers A p of a matrix A∈ℂ n×n. The algorithm starts with a Schur decomposition, takes k square roots of the triangular factor T, evaluates an [m/m] Padé approximant of (1-x)p at I-T 1/2 k, and squares the result k times. The parameters k and m are chosen to minimize the cost subject to achieving double precision accuracy in the evaluation of the Padé approximant, making use of a result that bounds the error in the matrix Padé approximant by the error in the scalar Padé approximant with argument the norm of the matrix. The Padé approximant is evaluated from the continued fraction representation in bottom-up fashion, which is shown to be numerically stable. In the squaring phase the diagonal and first superdiagonal are computed from explicit formulae for T p/2 j, yielding increased accuracy. Since the basic algorithm is designed for p∈(-1,1), a criterion for reducing an arbitrary real p to this range is developed, making use of bounds for the condition number of the A p problem. How best to compute A k for a negative integer k is also investigated. In numerical experiments the new algorithm is found to be superior in accuracy and stability to several alternatives, including the use of an eigendecomposition and approaches based on the formula A p=exp(p log(A)). Multiprecision Algorithms for Computing the Matrix Logarithm Massimiliano Fasi , Nicholas J. Higham Abstract Two algorithms are developed for computing the matrix logarithm in floating point arithmetic of any specified precision. The backward error-based approach used in the state of the art inverse scaling and squaring algorithms does not conveniently extend to a multiprecision environment, so instead we choose algorithmic parameters based on a forward error bound. We derive a new forward error bound for Padé approximants that for highly nonnormal matrices can be much smaller than the classical bound of Kenney and Laub. One of our algorithms exploits a Schur decomposition while the other is transformation-free and uses only the computational kernels of matrix multiplication and the solution of multiple right-hand side linear systems. For double precision computations the algorithms are competitive with the state of the art algorithm of Al-Mohy, Higham, and Relton implemented in logm in MATLAB. They are intended for computing environments providing multiprecision floating point arithmetic, such as Julia, MATLAB via the Symbolic Math Toolbox or the Multiprecision Computing Toolbox, or Python with the mpmath or SymPy package. We show experimentally that the algorithms behave in a forward stable manner over a wide range of precisions, unlike existing alternatives. Download PDF Figures Tables Close figure viewer Back to article Figure title goes here Change zoom level Go to figure location within the article Download figure Toggle share panel Share on social media Toggle information panel All figures All tables xrefBack.goTo xrefBack.goTo Request permissions Expand All Collapse Expand Table Show all references SHOW ALL BOOKS Authors Info & Affiliations HomeSIAM Journal on Matrix Analysis and ApplicationsVol. 22, Iss. 4 (2001)10.1137/S0895479800368688 Share ###### PREVIOUS ARTICLE Approximating the Logarithm of a Matrix to Specified Accuracy Previous###### NEXT ARTICLE Joint Approximate Diagonalization of Positive Definite Hermitian Matrices Next back Previous articleNext article Society for Industrial and Applied Mathematics Society for Industrial and Applied Mathematics 3600 Market Street, 6th Floor Philadelphia, PA 19104 USA © 2025 Society for Industrial and Applied Mathematics Browse ### Browse Journals E-books Bookstore Proceedings Alerts ### Alerts Sign up/Manage Email Alerts Information ### Information For Journal Authors For Book Authors For Librarians Help Terms of Use & Privacy Policy Accessibility Statement About ### About SIAM Join SIAM Donate to SIAM Request Username Can't sign in? 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https://www.learntheta.com/maths-application-of-herons-formula-area-of-quadrilaterals/
Skip to content Application of Heron’s Formula: Area of Quadrilaterals Finding the area of a quadrilateral can be simplified by dividing it into two triangles. Since the area of a triangle is relatively easy to calculate (using the formula involving base and height or Heron’s formula), we can leverage this to find the area of more complex shapes like quadrilaterals. The process involves drawing a diagonal across the quadrilateral, which effectively splits it into two triangles. Then, calculate the area of each triangle separately and add them to obtain the total area of the quadrilateral. Formulae The primary formulae used here are: Area of a triangle (base and height): Area=12×base×height Area of a triangle (using Heron’s formula – when all three sides are known): Area=√s(s−a)(s−b)(s−c), where s is the semi-perimeter (s=a+b+c2), and a, b, and c are the side lengths. Area of a quadrilateral: Areaquadrilateral=Areatriangle1+Areatriangle2 Examples Example-1: Find the area of a quadrilateral ABCD where AC = 10 cm, the perpendicular distance from B to AC is 4 cm, and the perpendicular distance from D to AC is 6 cm. Solution: Divide the quadrilateral into two triangles: △ABC and △ADC. Area of △ABC=12×AC×(perpendicular from B to AC)=12×10×4=20 cm2. Area of △ADC=12×AC×(perpendicular from D to AC)=12×10×6=30 cm2. Area of quadrilateral ABCD = Area of △ABC + Area of △ADC = 20+30=50 cm2. Example-2: A quadrilateral has vertices A(1,1), B(4,5), C(7,1) and D(4,-3). Find its area. Solution: Draw a diagonal, say AC. The coordinates of A and C are (1,1) and (7,1), respectively. The length AC = 7-1 = 6 units. The perpendicular distance from B to AC is the y-coordinate difference, 5-1=4. Therefore, the area of △ABC=(1/2)∗6∗4=12 square units. The perpendicular distance from D to AC is the y-coordinate difference, 1-(-3)=4. Therefore, the area of △ADC=(1/2)∗6∗(−3−1)=12 square units. The area of quadrilateral ABCD = Area of △ABC + Area of △ADC = 12+12=24 units2. Theorem with Proof Theorem: The area of a quadrilateral can be found by dividing it into two triangles and summing their areas. Proof: Let’s consider a quadrilateral ABCD. Draw a diagonal AC, dividing the quadrilateral into two triangles, △ABC and △ADC. The area of △ABC is given by Area△ABC=12×baseAC×heightB. (Let hB be the height from B to AC) The area of △ADC is given by Area△ADC=12×baseAC×heightD. (Let hD be the height from D to AC) The area of the quadrilateral ABCD is the sum of the areas of these two triangles: AreaABCD=Area△ABC+Area△ADC AreaABCD=(12×baseAC×hB)+(12×baseAC×hD) AreaABCD=12×baseAC×(hB+hD) This proves that dividing a quadrilateral into two triangles allows us to calculate its area by summing the areas of the individual triangles. Therefore, area of quadrilateral is simply the addition of areas of triangles. Common mistakes by students Common mistakes include: Incorrectly identifying the base and height of the triangles. Forgetting to divide by 2 when calculating the area of a triangle. Confusing the lengths of the sides of the triangles with the height. Not correctly understanding the co-ordinate system to calculate height of the triangles. Real Life Application This concept is used in various real-life situations, such as: Land Surveying: Surveyors often divide irregular land plots (which can be considered as quadrilaterals) into triangles to calculate their areas and determine property boundaries. Architecture and Construction: Architects and builders use this method to calculate the area of floors, walls, and roofs of buildings. They might divide complex shapes into simpler ones, like quadrilaterals, and then into triangles. Computer Graphics: In computer graphics, complex shapes are often rendered using triangles (triangulation), and the area calculation is crucial for rendering and other calculations. Navigation: Calculating the areas of regions on maps to estimate the area of territories or plan routes. Fun Fact Did you know that any polygon can be broken down into triangles? This is a fundamental concept in geometry, and it’s why triangles are so important in various fields like computer graphics and engineering. You can always decompose a polygon into triangles, making area calculations and other geometric analysis much easier. Recommended YouTube Videos for Deeper Understanding Practice Questions Q.1 A quadrilateral ABCD has vertices A(1,1), B(4,2), C(5,5), and D(2,4). What is the area of quadrilateral ABCD? Check Solution Ans: A Divide the quadrilateral into two triangles, △ABC and △ADC. Area of △ABC=12|(1(2−5)+4(5−1)+5(1−2))|=12|−3+16−5|=12|8|=4. Area of △ADC=12|(1(5−4)+5(4−1)+2(1−5))|=12|1+15−8|=12|8|=4. Area of quadrilateral ABCD = Area of △ABC + Area of △ADC = 4+4=8. Q.2 A quadrilateral has vertices at (0,0), (4,0), (5,3), and (1,3). Find the area of the quadrilateral. Check Solution Ans: A Divide the quadrilateral into two triangles, △ABC and △ADC. Let the points be A(0,0), B(4,0), C(5,3) and D(1,3). Area of △ABC=12|0(0−3)+4(3−0)+5(0−0)|=12|12|=6. Area of △ADC=12|0(3−3)+5(3−0)+1(0−3)|=12|15−3|=12|12|=6. Area of quadrilateral = 6+6=12. Q.3 The lengths of the diagonals of a rhombus are 10 cm and 24 cm. What is the area of the rhombus? Check Solution Ans: A The area of a rhombus can be found by 12d1d2, where d1 and d2 are the lengths of the diagonals. Area = 12×10×24=120 square cm. Q.4 A quadrilateral PQRS has diagonals PR and QS. The length of PR is 14 cm, and the perpendicular distance from Q to PR is 6 cm, and the perpendicular distance from S to PR is 8 cm. What is the area of quadrilateral PQRS? Check Solution Ans: A The area of quadrilateral PQRS is the sum of the areas of the two triangles, △PQR and △PSR. Area of △PQR=12×PR×h1=12×14×6=42. Area of △PSR=12×PR×h2=12×14×8=56. Area of quadrilateral PQRS = 42+56=98. Q.5 If the sides of a quadrilateral are tangent to a circle, and two opposite sides are of length 10 cm and 14 cm, what is the perimeter of the quadrilateral? Check Solution Ans: A A quadrilateral circumscribing a circle has the sum of opposite sides equal. Let the lengths of the sides be a,b,c,d. We are given that a+c=b+d. Let the sides be a=10 and c=14. Then b+d=10+14=24. Therefore the perimeter is a+b+c+d=(a+c)+(b+d)=10+14+24=2(10+14)=48. Next Topic: Cube and Cuboid: Surface Area & Volume Practice Exta Questions for Class 9 Maths Improve Maths with LearnTheta’s AI Practice Adaptive Practice | Real Time Insights | Resume your Progress Start Your Free Trial Today! Try a better way to Practice? Yes No Scroll to Top Start AI-powered Practice! 🚀
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https://www.youtube.com/watch?v=garegCgMxxg
Angular motion variables | Moments, torque, and angular momentum | Physics | Khan Academy Khan Academy Physics 134000 subscribers 1510 likes Description 290385 views Posted: 29 Jul 2016 In this video David explains the meaning of angular displacement, angular velocity, and angular acceleration. Watch the next lesson: Missed the previous lesson? Physics on Khan Academy: Physics is the study of the basic principles that govern the physical world around us. We'll start by looking at motion itself. Then, we'll learn about forces, momentum, energy, and other concepts in lots of different physical situations. To get the most out of physics, you'll need a solid understanding of algebra and a basic understanding of trigonometry. About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy’s Physics channel: Subscribe to Khan Academy: 47 comments Transcript: Introduction - [Instructor] I found that for many people the hardest part about solving a rotational motion problem is just keeping track of all the new names for all the rotational quantities that there are. So in this video, I want to go over all the different rotational variables like angular displacement, angular velocity, and angular acceleration. We'll explain what they mean, how they're defined, and how you can solve for them, so let's do this. Let's consider this example. Say take a tennis ball, you tie a string to it, and you whirl the tennis ball around in a circle. If you did this and you wanted to start defining motion variables that would describe the rotational motion of this tennis ball, maybe the most basic quantity you'd come up with would just be how much angle has this tennis ball swung through during its motion. So if we imagine the tennis ball starting there and it rotates over to here, we could define a quantity that just says how much angle has this thing gone through. Angular displacement And that would be what's called the angular displacement. And it's given the symbol delta theta, because theta is the angle and delta theta is the change in the angle, so this is really theta final minus theta initial. For instance if we started the tennis ball over here at zero and we ended it at 180, theta final would be 180, theta initial would be zero, so our angular displacement would be 180 degrees or pie radians. And if we started at zero and went through an entire circle all the way, and then another circle all the way, our angular displacement wouldn't be zero. It would technically be two whole revolutions, which would be either 720 degrees or four pie radians. And we don't even have to start at the zero. Our theta initial could be over here at 180, and we'd go down to 270, in which case the angular displacement would be 90 degrees or pie over two radians. Angular velocity So this is how we define the angular displacement and we typically measure it in radians, as opposed to degrees for reasons that I'll show you in just a second. And the name for this symbol here is theta. And we should mention that this is analogous to how we defined the regular displacement, so if you imagine a tennis ball just going in a straight line, the regular displacement was a defined b, the final position minus the initial positions, which we called delta x. And that's just usually called the displacement, which is measured in meters. Okay, so now we know how to quantify the amount of angle that this ball has rotated through, but another quantity that might be useful is the rate at which it's traveling through that angle. Just like up here, knowing about the displacement is good, but you might want to know about the rate that it's being displaced. In terms of regular linear quantities that was called the velocity of the ball, and it was defined to be the displacement per time. So down here we'll define a similar quantity, but it's going to be the angular velocity, which is defined analogously to the regular velocity. If regular velocity is displacement per time, the angular velocity is going to be the angular displacement per time. And the symbol we used to represent angular velocity is the Greek letter omega, which looks like a w, but it's really the Greek letter omega. And the units of omega, angular velocity, are going to be radians per seconds. Since delta theta, the angular displacement is in radians, and the time is in seconds. Just like how regular velocity had units of meters per second, angular velocity has units of radians per second. What is angular velocity mean? What is this omega? It represents the rate at which an object is changing its angle in time. So let's say the tennis ball starts here, and it's going through a circle at this leisurely rate, that means the rate at which it's changing its angle is very small and it has a very small omega. Whereas if you had this tennis ball going through a circle very fast, the rate at which it's going in a circle would be large and that means the angular velocity and omega would also be large. Angular acceleration So the velocity and the angular velocity are related, they're not equal because the velocity gives you how many meters per second something is going through, and the angular velocity gives you how many radians per second it's going through, but if it's got a larger angular velocity, it's going to have a larger velocity as well. And just like velocity is a vector, angular velocity is also a vector, so I'll put an arrow over this omega. Which way does it point? Technically speaking, you'd use the same right hand rule you use to determine the direction of the angular displacement. But again if it's rotating counter clockwise, we can just consider that to be positive, and if it's rotating clockwise, we can consider that to be a negative omega, or a negative angular velocity. So let me get rid of these, and let's define our last angular motion variable. You can probably guess what it is. There's regular displacement and there's angular displacement. There's regular velocity and there's angular velocity. And then the next logical step in this motion variable sequence would be the acceleration, which was defined for regular variables to be the change in velocity over the change in time. So we'll define an analogous angular quantity that would be the angular acceleration. And it's going to be defined to be, instead of change in velocity over change in time, it's going to be the change in the angular velocity over the change in time. And the letter we use to denote angular acceleration is this Greek letter alpha, so this is the Greek letter alpha. It looks like a little fishy, and that represents the angular acceleration of an object. So what does this angular acceleration mean? Well, looking at the units, helps us to figure this out, so the units of regular acceleration were meters per second per second, so regular acceleration represented the rate at which the velocity is changing, and that's the same analogous definition down here. The units down here are going to be radians per second per second, so this is going to represent this angular acceleration is going to represent the rate at which the angular velocity is changing. Angular acceleration examples What would that look like? Well if we've got this ball rotating in a circle, if it's rotating at a constant rate, there's no angular acceleration since the omega, the angular velocity wouldn't be changing. So in other words, if it's rotating at a constant rate, there's no change in the angular velocity, and that means there's no angular acceleration. But conversely, if it starts off moving slowly, and it speeds up its angular velocity is increasing, then there is an angular acceleration because there's a change in the angular velocity of this ball. And just like any acceleration, this angular acceleration can increase the angular velocity and speed something up. Or it can slow the object down and decrease the angular velocity. But if the angular velocity is remaining constant, in other words it's rotating in a circle at a constant rate, then the angular acceleration is zero and that means alpha equals zero. Angular acceleration vs angular velocity And just like the rest of these motion variables, angular acceleration is a vector, just like regular acceleration is a vector. And the direction that the angular acceleration points will be in the direction of the change in the angular velocity. So in other words, if this tennis ball is speeding up, then the angular acceleration is pointed in the same direction as the angular velocity. And if the angular velocity is slowing down, the angular acceleration points in the opposite direction to the angular velocity. At this point, I wouldn't blame you if you weren't like why, why do we need to define all these new angular variables when we already had all these regular variables up here. And the answer is that it's the same reason we define most variables in physics, because it turns out to be really convenient to do so, and these angular variables are going to be way more convenient to describe an object that's rotating than these regular variables. For this reason, imagine you wanted to describe not just the ball on the end of the string, but all points on the string as well. If you limited yourself to only these regular motion variables, you'd run into a problem. You'd realize that his ball goes through a circle in a certain amount of time, but every point on this string also goes through a circumference in that same amount of time, so the velocity of the ball is going to be greater than the velocity of points on the string that are closer to the center. Because everything's taking the same amount of time to go through one circle, but the circle the ball goes through has a larger circumference than the circle that points nearer to the center do. And so all points on this string are going to have a different velocity the closer you get to the center of the string. So trying to describe its motion with just velocity might be a nightmare, whereas if you were just going to use angular velocity, note that every point on the string, including the ball moved through the same amount of angle in the same amount of time. They don't move through the same amount of meters per second, but they do move through the same amount of radians per second because when the ball has rotated through two pie radians, once full circle, every point on this string has rotated through two pie radians. If this ball and string are going to maintain the same shape. Sample problem So that's the great thing about these angular motion variables, every point on a rigid object is going to have the same angular displacement, the same angular velocity, and the same angular acceleration. It won't matter what point you're talking about. The angular displacement, angular velocity, and angular acceleration will be the same for every point on that rotating object. Alright, so before this gets too abstract, let's try a sample problem. Let's say that the ball starts over here at rest, and it rotates all the way to this point in four seconds. So it started over here at rest, and it took four seconds for it to rotate over to this point. And let's say when the ball makes it over to this side, it's going 1.57 radians per second. Let's say that's the final angular velocity. So let's just go through and try to figure these out. What would the angular displacement be for this example? Well if the ball started here and it made it over to here, the angular displacement would be pie radians, or 180 degrees. What would the angular velocity be? Well it started at rest, so initially omega was zero at this point here, and then finally it tells us what the final omega would be, 1.57. So you might wonder, what would we do with this formula? What if we just used this formula, what would we get? Well if we used that formula there, we're going to get that it went pie radians, and it did it in four seconds, which gives us 0.785 radians per second. So you might be like wait a minute, this omega doesn't correspond to the initial omega or the final omega. What is this corresponding to? Well this would be the average omega. This is the average angular velocity between this initial and final point. This at rest initially shows that the omega started off as zero. The instantaneous omega was zero, and the instantaneous omega, or the final angular velocity would be 1.57, so you got to be careful. The instantaneous values are not necessarily equal to the average value. You can get the average value by taking the change in theta over the change in time, but it doesn't necessarily give you the instantaneous angular velocity at a specific point on the trip. And we can find the angular acceleration as well if we use this formula. The change in omega over the change in time. That would be the angular acceleration, our omega final minus omega initial over time would come out to be 1.57 as our final angular velocity minus zero was our initial angular velocity, and that took four seconds to accomplish, so our angular acceleration would come out to be 0.393 radians per second per second, or you can write that as radians per second squared. Now technically that is also the average angular acceleration during this trip, but if the angular acceleration was constant during this trip, which in almost all cases we're going to look at, the angular acceleration is going to be constant. If that's the case, this would be both the average value and the instantaneous value of the angular acceleration at every point on the trip since the angular acceleration would be remaining constant. So in this example, we can say that the angular displacement was pie radians. The average angular velocity was 0.785 radians per second. The initial angular velocity was zero. The final angular velocity was 1.57, and the angular acceleration was 0.393 radians per second squared. So recapping, the angular displacement represents the angle through which an object is rotated. It's typically measured in radians, and it's represented with a delta theta. The angular velocity represents the rate at which an object is rotating. It's measured in radians per second, and it's represented with a Greek letter omega. And the angular acceleration represents the rate at which an object is changing its angular velocity, so if an object rotates at a constant rate, there is zero angular acceleration, but conversely, if an object's rotation is speeding up or slowing down, there must be angular acceleration. It's measured in units of radians per second per second or radians per second squared, and it's represented with the Greek letter alpha.
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https://pmc.ncbi.nlm.nih.gov/articles/PMC7849738/
Basic physiology of the blood-brain barrier in health and disease: a brief overview - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer | PMC Copyright Notice Tissue Barriers . 2020 Nov 15;9(1):1840913. doi: 10.1080/21688370.2020.1840913 Search in PMC Search in PubMed View in NLM Catalog Add to search Basic physiology of the blood-brain barrier in health and disease: a brief overview Mehmet Kaya Mehmet Kaya a Koç University School of Medicine Department of Physiology, Koç University Research Center for Translational Medicine, Istanbul, Turkey Find articles by Mehmet Kaya a,✉, Bulent Ahishali Bulent Ahishali b Koç University School of Medicine Department of Histology and Embryology, Koç University Research Center for Translational Medicine, Istanbul, Turkey Find articles by Bulent Ahishali b Author information Article notes Copyright and License information a Koç University School of Medicine Department of Physiology, Koç University Research Center for Translational Medicine, Istanbul, Turkey b Koç University School of Medicine Department of Histology and Embryology, Koç University Research Center for Translational Medicine, Istanbul, Turkey ✉ CONTACT Mehmet Kaya mehmetkaya@ku.edu.tr Koç University School of Medicine Department of Physiology, Koç University Research Center for Translational Medicine, Istanbul, 34450, Turkey Collection date 2021. © 2020 Taylor & Francis Group, LLC PMC Copyright notice PMCID: PMC7849738 PMID: 33190576 ABSTRACT The blood-brain barrier (BBB), a dynamic interface between blood and brain constituted mainly by endothelial cells of brain microvessels, robustly restricts the entry of potentially harmful blood-sourced substances and cells into the brain, however, many therapeutically active agents concurrently cannot gain access into the brain at effective doses in the presence of an intact barrier. On the other hand, breakdown of BBB integrity may involve in the pathogenesis of various neurodegenerative diseases. Besides, certain diseases/disorders such as Alzheimer’s disease, hypertension, and epilepsy are associated with varying degrees of BBB disruption. In this review, we aim to highlight the current knowledge on the cellular and molecular composition of the BBB with special emphasis on the major transport pathways across the barrier type endothelial cells. We further provide a discussion on the innovative brain drug delivery strategies in which the obstacle formed by BBB interferes with effective pharmacological treatment of neurodegenerative diseases/disorders. KEYWORDS: Blood-brain barrier, endothelial cells, tight junctions, paracellular permeability, transcellular permeability Introduction The brain is the most critical organ that controls body systems in humans. Oxygen and nutrients, mainly glucose and amino acids, are supplied to the cells in the brain parenchyma by an elaborate network of blood capillaries. The estimated total length of brain microvessels is about 600–700 km and the total area of the endothelial surface in brain vasculature including capillaries, venules, arterioles, veins, and arteries approximates 20 m 2.1,2 The brain is extremely sensitive to a wide range of potentially toxic substances in circulation, and the proper neuronal function necessitates an optimal microenvironment that is controlled and regulated by three different barrier systems; the blood-brain barrier (BBB) formed by brain microvessel endothelial cells (Figure 1), the blood-cerebrospinal fluid barrier (BCSFB) formed by choroid plexus epithelial cells (Figure 2Figure 3), and the meningeal barrier formed by arachnoid epithelial cells.3–5 It is suggested that neuronal homeostasis within the brain parenchyma is mainly regulated by the BBB since the total area of the luminal surface with BBB activity is estimated to be about 1000 times larger than that with BCSFB.6 Figure 1. Open in a new tab Schematic drawing of the BBB constituted by barrier type endothelial cell with TJ sharing the basement membrane with pericyte and the surrounding astrocyte endfeet. Created with BioRender.com Figure 2. Open in a new tab Schematic drawing of the BCSFB constituted by epithelial cells of choroid plexus with TJs, which secrete CSF derived from plasma in blood capillary without barrier properties into the ventricular space lined with ependymal cells. Created with BioRender.com Figure 3. Open in a new tab An electron micrograph from our image archive showing a pericyte (p) partly investing the endothelial cells of blood capillaries. Note that both endothelial cells and the pericyte are embedded in the basement membrane marked by extravasated electron-dense horseradish peroxidase tracer accumulation (arrows) owing to BBB disruption. A: astrocyte endfoot The concept of the BBB was first established in the late 19 th century by Paul Ehrlich, who observed that trypan blue dye injected into the rat circulation resulted in the staining of peripheral organs but not the brain and spinal cord. In the following years, Goldman injected trypan blue into the cerebrospinal fluid (CSF), and he demonstrated that the staining was only restricted to the central nervous system (CNS), but not to the other body tissues.7 By the advent of electron microscopy and its widespread use in the evaluation of biological tissues in the 1960s, the presence of the BBB was confirmed using an electron-dense tracer, horseradish peroxidase (40 kD), which had been observed to pass through the vascular endothelium in peripheral tissues in contrast to that in the brain.8 The BBB, as a dynamic regulatory interface between blood and brain, protects neuronal microenvironment required for the proper functioning of neuronal circuits, synaptic transmission and remodeling, angiogenesis, and neurogenesis, by constantly controlling trafficking of molecules and preventing circulatory immune cell entry into the brain via paracellular and transcellular pathways.3,9–12 The access of certain blood-borne neuroactive solutes, such as glutamate, glycine, norepinephrine, epinephrine, and peptide hormones into the brain is also significantly limited by the action BBB.13–15 A healthy BBB not only protects the neurons but also is crucial for the physiologic functions of glial cells and pericytes. The afore-mentioned protective activities of BBB brings with a concomitant obstacle for the access of therapeutic agents into the brain at effective doses for the treatment of neurodegenerative diseases. The presence of an intact BBB excludes approximately 98% of small molecule drugs and nearly all large therapeutics, such as recombinant peptides, proteins, anti-sense-agents, and genetic vectors from the brain.16 Moreover, the fraction of therapeutic antibodies such as immunoglobulin G that reaches into the brain parenchyma following intravenous administration is estimated to be as low as 0.1%.17 On the other hand, certain systemic diseases or CNS disorders are likely to evoke alterations in BBB integrity, leading to BBB disruption and loss of neuronal homeostasis.18–21 In this review, we present an overview of the structure and function of the BBB in both healthy and pathological conditions, the alterations in BBB integrity associated with neurodegenerative disorders/diseases and novel strategies to enhance targeted drug delivery into the brain. Neurovascular unit In the last 2 decades, the term “neurovascular unit” has started to be used to define a well-structured complex that is involved in the development and maintenance of BBB integrity and in the regulation of cerebral blood flow. The major constituents of the unit are barrier type endothelial cells interacting with the basement membrane, pericytes, vascular smooth muscle cells, astrocytes, microglia, oligodendroglia, and neurons.22–26 Barrier type endothelial cells The cellular structures that lie in the interface between the blood and the brain are endothelial cells, pericytes, and astrocyte endfeet, which along with a cellular product, basement membrane, collectively compose the major constituents of the BBB and orchestrate the trafficking of molecules and cells between the two compartments. The primary part of the BBB is made up of capillaries that have luminal diameters of less than 10 µm.27,28 In contrast to the various types of peripheral capillaries with different capacities of permeability found in other organs, brain capillaries display barrier type endothelial cells with a variety of distinguished properties, which render them the chief element of BBB structure and the main actor in the maintenance of neuronal homeostasis. Accumulated data show that brain capillary endothelial cells are sealed with tight junctions (TJs), which restrict the paracellular transport, and lack endothelial fenestrations and possess relatively few caveolar vesicles which limit the transcellular transport.3,29,30 In addition, barrier type endothelial cells exhibit specific transport and carrier proteins that are located both in the luminal and abluminal plasma membranes.31–33 Moreover, these cells are also considered to form an endocrine secretory tissue that produces a variety of local hormones such as nitric oxide, prostaglandins, and cytokines.34–36 A continuous basement membrane, pericytes and astrocyte endfeet surrounding the endothelium provide anatomical support to the BBB.3,37,38 Physical barrier (Tight junctions) Barrier type endothelial cells are normally adjoined to each other by TJs and adherens junctions, which collectively constitute a physical barrier that limits the paracellular pathway.22 The exchange of polar substances between the blood and the brain is strictly controlled and significantly reduced by the TJ proteins.3,39 Metabolic barrier The brain capillary endothelial cells express a variety of enzymes including γ-glutamyl transpeptidase, alkaline phosphatase, aromatic acid decarboxylase, monoamine oxidase, and cytochrome P450, which metabolize and inactivate molecules such as neuroactive and neurotoxic compounds, peptides, and ATP.6,36,40–43 Barrier type endothelial cells also possess certain crucial transporters such as Na+-K+ATPase, and glucose transporter (Glut)-1 which take part in the regulation of the composition of neuronal microenvironment.44,45 Efflux barrier The barrier type endothelial cells of the brain capillaries express certain efflux transporters, designated as multidrug-resistance transporters localized predominantly on the luminal plasma membrane.46,47 These transporters include P-gp, also called mdr-1, and breast cancer resistance protein which extrude the administered xenobiotics, compounds that include lipophilic and cationic drugs, from the brain capillary endothelial cells back to the circulation and hence reduce the delivery of drugs into the brain parenchyma at effective doses thereby posing an obstacle to the treatment of neurodegenerative disorders/diseases. A number of chemical compounds and chemotherapeutic agents used in daily clinical practice have been described as the substrates of these transporters.48,49 Successful management of tumors and refractory epilepsy is reported to be overwhelmed by the activity of P-gp.50,51 Although inhibitors of P-gp and breast cancer resistance protein have been used effectively to overcome the drug resistance in experimental animals, human data are still lacking.52–54 Basement membrane The basement membrane of the capillary wall in the brain is structurally an organized protein sheet with a thickness of 50–100 nm and surrounds both endothelial cells and pericytes. It is a highly dynamic constituent of the BBB and plays an essential role in the maintenance of BBB integrity.55–57 The contents of the basement membrane are secreted by endothelial cells and pericytes and are mainly composed of laminins, collagen type IV isoforms, fibrillins, vitronectin, fibronectin, elastin, nidogens, and heparan sulfate.56,58 In addition, soluble factors (e.g., growth factors and cytokines), enzymes responsible for matrix degradation, and proteins such as lectins and semaphorins are also present in the basement membrane structure.56,59,60 Laminins primarily play a role in the organization and scaffolding, and collagen type IV is essential for the stability of the basement membrane.59,61 Matrix metalloproteinases that are activated by certain pathological insults may disrupt the integrity of the basement membrane resulting in BBB breakdown through an impairment in the functional activity of TJ proteins.19,22 Pericytes Pericytes, one of the components of the neurovascular unit, are located between endothelial cells and astrocyte endfeet (Figure 1). They are embedded in the same basement membrane surrounding the endothelial cells and thus are physically separated from both endothelial cells and astrocyte endfeet.3,62,63 The available data on the percentage of the surface area of the abluminal plasma membrane of endothelial cells covered by the pericytes are contradictory with reported values ranging from 22 to 99% .64–68 Pericytes are contractile cells, which provide physical support to and determine the vasodynamic properties of brain capillaries and hence contribute to the regulation of cerebral blood flow by controlling the luminal diameter.69–71 A recent study suggested that pericytes can construct tunneling nanotubes that regulate neurovascular coupling and control capillary blood flow.72 Pericytes are also essential for the induction of barrier characteristics in the endothelial cells including the formation of TJs and are involved in the regulation of BBB integrity and the transport of substances into the brain parenchyma.62,63,73–76 Moreover, pericytes produce extracellular matrix proteins and play a crucial role in the regulation of endothelial cell proliferation, migration, and differentiation.77–80 Besides, they also take part in the clearance of toxic cellular byproducts.66,73,74,81 In experimental animals, pericyte deficiency induced by a platelet-derived growth factor mutation has been shown to cause a reduction in the expression of certain TJ proteins leading to a substantial BBB disruption.63,73 Astrocytes Astrocytes, the most abundant cell type in the brain, are characterized by their expression of the intermediate filament glial fibrillary acidic protein (GFAP) and their numerous cellular processes extending from the cell body. Most of these processes terminate as endfeet, which contact with the abluminal side of the basement membrane of brain capillaries to interact with endothelial cells and pericytes. Astrocytic perivascular endfeet are estimated to cover over 99% of the brain microvasculature wall (Figures 1 and 4).23,42,65 Figure 4. Open in a new tab A light micrograph from our image archive showing astrocytes labeled by immunostaining for GFAP in the hippocampal region of the brain. Note the microvessels almost entirely surrounded with astrocyte endfeet (arrows) As a component of the neurovascular unit, astrocytes contribute to the regulation of vascular tone and local blood flow into the brain parenchyma and hence play an important role in the transport of oxygen and nutrients to neurons to maintain brain homeostasis. Astrocytes are the main actors that determine neuronal activity by regulating ion concentration and extracellular pH within the interstitial space and the uptake of glutamate and GABA in the synaptic region.82–84 They express aquaporin-4 water channel proteins that promote perivascular clearance of waste material and hence form the newly characterized “glymphatic system” (CNS waste clearance system). Astrocytes also express certain transporters such as P-gp and Glut-1 along with Kir4.1 K+channel proteins that aid in the maintenance of the neuronal resting membrane potential by removing extracellular K +.85–89 Astrocytes orchestrate the development of BBB properties and barrier maturation by releasing specific factors .19,42,90–92 They are involved in the maintenance of BBB integrity by providing functional and anatomical support.93,94 The alterations in astrocyte characteristics are reported to be associated with impairment in BBB integrity.95 Data from the studies of our research group have shown BBB disruption along with alteration in GFAP immunoreactivity in astrocytes in experimental models of hypertension, febrile seizures, and irradiation.96–98 On the contrary, there are some reports in the literature that oppose the necessity of glial cells for the maintenance of BBB integrity.99,100 Furthermore, reactive astrocytes were found to disrupt the BBB integrity by releasing vascular endothelial growth factor.101 Microglia Microglia are long-living resident immune cells of the CNS and account for around 12–16% of the total cell population in the brain. They constitute the major cell type that acts in the protection of the brain against immunologic insults and thus contribute to the maintenance of neuronal homeostasis.102,103 The breakdown of BBB may alter microglial activity through interaction with activated endothelial cells even in the absence of neurodegeneration, and hence brain regions with hypertrophied/activated microglial-like cells associated with vasculature may potentially display vascular damage and BBB compromise.104 The activated microglia produce pro-inflammatory cytokines, including IL-1β and TNF-α, that further enhance the degree of BBB disruption.105,106 Activation of microglia by lipopolysaccharide has been shown to decrease trans-endothelial electrical resistance (TEER) by disrupting TJ proteins, including claudin-5 and zonula occludens (ZO)-1 in an in vitro model of BBB.107 Although microglia are defined as the cells that form the first line of defense against immunologic compromise, a putative role of these cells with regard to the maintenance of BBB integrity in physiologic conditions or restoration of disrupted BBB in the course of CNS diseases/disorders remains unclear. Neuron Almost every neuron in the human brain is estimated to be nourished by a capillary microvessel positioned within an average distance of 15 μm.108,109 Accumulated data have demonstrated that there is a direct interaction between neurons and barrier type endothelial cells, pericytes, as well as astrocytes.22,23,110,111 The soma, axon, or dendrite of a neuron in close proximity to a brain capillary may contact pericytes and endothelial cells via the basement membrane.6,112 Communication of neurons with glial and endothelial cells is essential for their survival and functions. Depending on metabolic requirements, neurons not only modulate various endothelial cell functions, including permeability, by activating specific enzymes expressed by the endothelial cells but also release growth factors to stimulate angiogenesis.19,113,114 In the meantime, alterations in neuronal activity have been reported to affect BBB integrity.115 Trans-endothelial electrical resistance and expression of TJ proteins have been found to be positively influenced when endothelial cells were co-cultured with neurons in vitro.116 Functions of the blood-brain barrier Brain capillaries, like their peripheral counterparts, essentially provide oxygen and nutrients to parenchymal cells, including neurons, and remove the produced waste materials. Importantly, endothelial cells of the microvessels in the brain display strong barrier properties which enable strict control of ionic and fluid movements between the circulation and the brain parenchyma to regulate the neuronal microenvironment. The transfer of substances from the blood to the brain parenchyma across the BBB is accomplished through a transcellular route in which receptor and carrier-mediated transporters are critical for transcytosis, and paracellular route in which TJs are the chief determinants of permeability. Characteristics of the blood-brain barrier The endothelial cells of the brain capillaries are the fundamental anatomical structures of the BBB. These cells express not only a variety of specific transport and carrier proteins which enable tightly controlled trafficking of molecules, but also TJ proteins which account for one of the barrier characteristics reflected by high electrical resistance (approximately 1800 Ω cm 2) compared with that in peripheral capillaries (2–20 ohm cm 2).117,118 Tight junctions The intercellular cleft between adjacent endothelial cells of brain capillaries houses two major types of junctional complex; TJs and adherence junctions (Figure 5Figure 6Figure 7). Tight junctions are highly dynamic structures that effectively limit the movement of water and solutes and regulate lateral diffusion through the paracellular pathway.22,23,119,120 These structures are formed by transmembrane proteins claudins, occludin, and junctional adhesion molecules (JAMs) which interact with the actin cytoskeleton of the endothelial cells by a number of cytoplasmic accessory proteins including ZO proteins, cingulin, AF-6, and 7H6.22,121–124 The interaction between ZO proteins and transmembrane proteins, including claudins and occludin have been shown to determine the stability and function of TJs.22,125–127 In contrast to the epithelial cells which exhibit intercellular gaps sealed by TJs localized at the most apical point of cellular attachment immediately above the clearly distinguishable adherens junctions, barrier type endothelial cells in the brain are joined together by TJs and adherens junctions showing more variable localizations and intermingled appearances.128 Figure 5. Open in a new tab Schematic drawing of the junctional complex between barrier type endothelial cells of the brain, which controls the trafficking of substances through the paracellular pathway. The major proteins comprising TJs and adherens junctions and their linkage to the actin cytoskeleton are illustrated. Created with BioRender.com Figure 6. Open in a new tab The schematic drawing of various routes of transcellular transport across barrier type endothelial cells in the brain. A: passive diffusion, B: efflux transport, C: carrier-mediated transport, D: receptor-mediated transport, E: adsorptive-mediated transport, F; cellular transport. Created with BioRender.com Figure 7. Open in a new tab An electron micrograph from a previous study from our research group (reproduced from Ref. # 96 with permission from Elsevier Science) showing a capillary from the hippocampus region of the brain of a rat with cortical dysplasia exposed to febrile seizures. Note a conglomerate of caveolar vesicles within a cargo (arrow in the inset) in the cytoplasm of a brain capillary endothelial cell and the intensive pericapillary edema with swollen astrocyte endfeet There are currently defined 27 members of claudins, and a number of them including claudin-1, −3, −5, −11, and −12 have been identified in the TJs between barrier type endothelial cells.19,129–132 However, only claudin-1, −3, and −5 are most likely responsible for controlling the paracellular pathway while the roles of claudin-11 and 12 with regard to barrier function are yet to be elucidated.23,130,133–135 Claudin-5 is the most enriched isoform in the brain endothelium and determines the sealing properties of TJs of the BBB.134 Claudin-5 knockout mice display a selective increase in paracellular permeability for small molecules.130 On the other hand, it is suggested that claudin-12 is not directly involved in the establishment or maintenance of BBB integrity.136 Occludin, the first TJ transmembrane protein described, has been defined as one of the major TJ proteins that controls the paracellular pathway of BBB.125,137 Early studies showed an enhancement in TEER and reduction in paracellular diffusion by an increase in the expression of the occludin protein.138,139 Moreover, dephosphorylation of occludin caused BBB failure in an experimental model of multiple sclerosis.140 An autopsy series of fatal human septic cases showed that occludin expression in the barrier type endothelial cells was lost in the brain.141 In diabetic rats, occludin and ZO-1 expression were decreased concomitant with an increase in BBB permeability to 14 C-sucrose.142 Data from the studies of our research group have shown BBB disruption along with a reduction of occludin immunoreactivity in barrier type endothelial cells in in vivo experimental models of sepsis, hypertension, and irradiation.97,98,143,144 In contrast, it is also suggested that occludin does not have the ability to establish TJ structure by itself but rather exerts a regulatory function on the barrier properties.145 Accordingly, occludin-deficient mice display well-developed TJ complexes without any evidence of BBB hyperpermeability.139 Among the three isoforms of JAMs, JAM-1, JAM-2, and JAM-3, an interaction with occludin and claudins in TJs of brain capillaries to provide cell-to-cell adhesion has been described for JAM-1.123,133,146 The members of JAM family also play an important role in leukocyte adhesion and transmigration to the brain parenchyma across BBB.147 Zonula occludens proteins (ZO-1, ZO-2, and ZO-3) are membrane-associated guanylate kinase homologs located in the cytoplasmic domain of TJs between endothelial cells. Among the members of ZO protein family, ZO-1 plays a central role in the assembly and organization of claudins, occludin, and JAMs, and links these TJ proteins to the cortical actin cytoskeleton.133,148–150 ZO-1 is also a central regulator of vascular endothelial-cadherin–dependent adherens junctions that orchestrate the tuning of cell-cell tension, migration, angiogenesis, and barrier formation.150 Data from our lab demonstrated decreased immunoreactivity for ZO-1 in brain capillary endothelial cells along with increased BBB permeability in an in vivo model of seizure and by radiation therapy in rats.97,151 Besides, the loss or dissociation of ZO-1 and occludin from the junctional complex associated with BBB disruption have been shown in a variety of pathological conditions such as hypoxia, subarachnoidal hemorrhage, and Parkinson’s disease.97,151–154 Adherens junctions Adherens junctions are constituted by Ca 1+-dependent transmembrane cadherin proteins that form homotypic adhesive complexes between neighboring endothelial cells and bind to the actin cytoskeleton via cytoplasmic anchoring proteins called catenins.3,155,156 Although the primary function of adherens junctions is the attachment of adjacent endothelial cells, they may also involve in the formation and maintenance of tightness of the TJs.119 However, the role of these junctional complexes in the development and normal physiology of BBB remains to be elucidated. Certain pathologic conditions causing disruption of adherens junctional proteins are associated with a loss of BBB integrity.123 Transport pathways across the blood-brain barrier Under physiological conditions, the BBB exhibits low permeability compared to peripheral blood vessels.157 Lipid solubility, electrical charge, molecular size, and hydrogen bonding capacity are the main determinants of the ability of a molecule in circulation to enter the brain across an intact functioning BBB. The paracellular passage of molecules is considerably minimized, while ions and solutes can diffuse between adjacent cells according to their concentration gradients. On the other hand, most of the nonpolar lipid-soluble molecules of small molecular weight (<400-500 Da) such as carbon dioxide, nitric oxide, ethanol, and oxygen, are readily transported into the brain parenchyma by the process of the passive diffusion mainly via transcellular route through the lipid bilayer of the endothelial cell membrane.3,12,16,42,158,159 Hydrophilic and charged molecules can only penetrate the BBB by active transport systems, including receptor-mediated transport, carrier-mediated transport, and adsorptive-mediated transport to enter the brain.42 Molecules such as glucose, transferrin, and amino acids and ions, including potassium, sodium, calcium, and bicarbonate, utilize these active transport systems; however, growth factors and cytokines have limited ability to permeate across the BBB.160,161 On the other hand, efflux transporters such as P-gp pump their substrates such as drugs and metabolites back to the circulation.162–164 The paracellular pathway The discovery of the ultrastructure of BBB enabled a clear understanding of the trafficking of substances across the barrier, and the paracellular pathway strictly controlled and regulated by the TJs localized along the interendothelial space was described.8 The specialized TJ proteins effectively prevent undesirable passive diffusion of lipophilic or low molecular weight substances and passage of immune cells through the gaps between the endothelial cells of brain capillaries and provide an obstacle for the bulk flow of water and plasma-sourced solutes by the paracellular route. The transendothelial pathway The trafficking of substances across the barrier type brain capillary endothelial cells is primarily mediated by the transcellular route. As a prominent characteristic of the barrier type endothelial cells, this pathway utilizes various types of influx transporters collectively called nutrient transporters, while efflux transporters pump their specific substrates back to the bloodstream .44,165,166 The transport of molecules by the transcellular pathway is bidirectional using receptor and carrier proteins located on both luminal and abluminal membranes of barrier type endothelial cells by energy-dependent or -independent processes.167 In addition, fluid-phase and adsorptive endocytosis is used to transport some non-lipid-soluble molecules of small molecular weight and macromolecules like albumin, immunoglobulins, and other proteins.168,169 Receptor-mediated transport The main pathway in the trafficking of molecules across the BBB is receptor-mediated transcytosis which mediates the transport of the circulatory substances including transferrin, low-density lipoproteins, leptin, insulin, and insulin-like growth factor into the brain; however, a shift to ligand-nonspecific caveolar transcytosis is observed by aging.33,41,170–173 Receptor-mediated transcytosis has recently been the focus of interest in targeted drug delivery studies in which molecular Trojan horses, vectors that can bind specific receptors of the pathway, are used to enable the transport of drugs into the brain at effective doses in certain CNS diseases/disorder resistant to pharmacological treatment.174,175 Carrier-mediated transport The circulatory-sourced vital substances that are essential for the energy and neurotransmitter metabolism in the brain, including nutrients such as glucose, vitamins, and hormones require the carrier-mediated transporters located both on luminal and abluminal plasma membranes of barrier type of endothelial cells to reach into the brain parenchyma through a saturable transport process.2,19,31,176–180 Glut-1 mediates the uptake of D-glucose, the main energy source of the brain, by barrier type endothelial cells and delivery to astrocytes and neurons.181–183 In the opposite direction, the uptake of D-glucose from the brain interstitium into the circulation is accomplished by the Na+-D-glucose cotransporter Sglt1, expressed in the brain capillary endothelial cells, which further contributes to the adjustment of glucose concentration in the brain interstitium.183 Glut-1 is also reported to be crucial for angiogenesis during brain development.184 The transport of certain ions in exchange of or simultaneously with other ions is exerted by exchanger pumps including sodium/potassium pump and sodium-hydrogen, chloride-bicarbonate and sodium-calcium exchangers, and cotransporters such as sodium-potassium-two chloride cotransporter localized on the abluminal and/or luminal side of the barrier type endothelial cells.185–187 Caveolae-mediated endocytosis Caveolae are characteristic flask-shaped membrane invaginations with a diameter of 50–80 nm, which are mainly responsible for endothelial transcytosis in barrier type of brain capillary endothelial cells.188,189 The caveolar membranes contain caveolin-1/2 and vesicle-associated membrane protein-2 as well as receptors for certain essential enzymes, hormones, plasma carrier proteins, and cytokines.190,191 Caveolin 1, which is the principal component of caveolae, can also influence the expression of TJ proteins.192 The expression of the caveolin-1 is significantly increased in barrier type endothelial cells under several pathological conditions and by aging.193–195 Major Facilitator Superfamily Domain containing 2a (Mfsd2a), a lipid transporter highly expressed in the endothelial cells of brain microvessels, inhibits caveolae production and hence plays an important role in barrier characteristics.196–198 Therefore, brain capillary endothelial cells exhibit few caveolae, whereas arteriolar endothelial cells in which Mfsd2a transcript levels are low display abundant caveolae.199,200 Knock-out of Mfsd2a in mice caused increased caveolae production and transcellular permeability in the brain microvasculature.195,197 Caveolae are characteristic flask-shaped membrane invaginations with a diameter of 50–80 nm, which contain caveolin-1/2 and vesicle-associated membrane protein-2 as well as receptors for certain essential enzymes, hormones, plasma carrier proteins, and cytokines.190,191 Caveolin 1, which is the principal component of caveolae, can also influence the expression of TJ proteins.192 Major Facilitator Superfamily Domain containing 2a (Mfsd2a), a lipid transporter highly expressed in the endothelial cells of brain microvessels, inhibits caveolae production and hence plays an important role in barrier characteristics.196–198 Therefore, brain capillary endothelial cells exhibit few caveolae, whereas arteriolar endothelial cells in which Mfsd2a transcript levels are low display abundant caveolae.199,200 Knock-out of Mfsd2a in mice caused increased caveolae production and transcellular permeability in the brain microvasculature.195,197 On the other hand, the expression of the caveolin-1 is significantly increased in barrier type endothelial cells under several pathological conditions and by aging.193–195 Increase of caveolar vesicles in brain capillary endothelium has been documented in experimental animals with hypertension and during febrile seizures in experimental models of cortical dysplasia, a malformation of the cerebral cortex .96,201 Efflux transport The multidrug-resistance proteins primarily consist of multidrug-resistance protein-1 or P-gp, multidrug resistance-associated protein, and breast cancer resistance protein.202 The P-gp, as an efflux pump, is located on the luminal side of the BBB and restricts the permeation of a large number of toxins into the brain parenchyma, providing neuroprotection and detoxification.203–205 On the other hand, the efflux pumps also restrict the entry of therapeutic agents, including antibiotics, chemotherapeutics, and antiepileptic drugs into the brain parenchyma at effective doses and thus severely contribute to the development of pharmacoresistance in the treatment of certain types of brain tumors and epilepsy.44,202,206 Cell movement across the BBB Under inflammatory conditions, circulatory-sourced immune cells and neutrophils can penetrate into the brain via paracellular pathway which requires opening and rearrangement of TJ complexes and transendothelial pathways which involves the dynamic organization of cellular processes of leukocytes called invadosomes and various vesicles and vesiculo-vacuolar organelles forming a transcellular pore through the endothelial cells.207–210 The cell surface levels of endothelial intercellular adhesion molecule-1 and caveolin-1 also play a crucial role in transcellular immune cell entry into the brain.211,212 Circumventricular organs In contrast to the capillaries located in the brain parenchyma, blood microvessels in the circumventricular organs do not display barrier properties. The endothelial cells of these microvessels have fenestrae, which allow the free diffusion of substances between the blood and CNS.37,213,214 These organs consist of secretory structures like pineal gland, subcommisural organ, median eminence, and choroid plexuses and sensory regions, including area postrema, subfornical organ, and organum vasculosum of the lamina terminalis.215 The exchange of hormones and other molecules between the circulation and CNS is accomplished mainly in circumventricular organs in which increased vascularization facilitates the sensory and secretory roles to mediate the communication between the brain and the periphery.215,216 BBB disruption in pathological conditions While the integrity of the BBB is crucial for the maintenance of neuronal homeostasis, alterations in functional and structural properties of the barrier are closely interrelated with the occurrence of certain brain pathologies. A variety of CNS diseases/disorders including epilepsy,175,217 ischemic stroke,218 multiple sclerosis,219 traumatic brain injury,220 and Alzheimer’s disease221 are characterized by BBB disruption. On the other hand, CNS manifestations associated with BBB breakdown may develop in systemic diseases such as sepsis222,223 and hypertension.201 Drug delivery into the brain for clinic implications Innovative strategies have been developed for overcoming BBB to enable the access of therapeutic drugs at effective doses in CNS diseases/disorders in experimental settings; however, their current use in clinical practice is still limited. Circumventing BBB by temporarily disrupting the TJs between brain capillary endothelial cells has been reported to allow the access of pharmacologic agents into the CNS in both experimental animals and humans. The intravenous administration of bradykinin and histamine, intraarterial infusion of hyperosmolar solutions like mannitol and application of transcranial focused ultrasound together with microbubbles temporarily open the TJs between barrier type endothelial cells in the brain.224–226 Moreover, conjugating the pharmacological agents with nanocarriers such as liposomes, nanopolymers, nanoparticles, viruses, and exosomes which normally have access into the brain using the endothelial transcellular transport mechanisms is being extensively studied in the last decade as an alternative approach to the drug delivery into the brain43,227–229 Future challenges Our understanding of the transport dynamics across BBB in both physiological and pathological conditions has advanced considerably in recent years by the rapid development of advanced molecular techniques, imaging modalities, and nanotechnology. In this context, the accumulated data on the behavior of BBB will pave the way to elucidate the mechanisms underlying the response of the neurovascular unit in neurodegenerative disorders/diseases and to develop novel therapeutic strategies. On the other hand, exploiting ways for the targeted delivery of pharmacologically active substances into the brain through transendothelial transport pathways using nanocarriers or by the reversible opening of both transcellular and paracellular routes will allow BBB permeation of therapeutic agents at effective doses. We believe that an elaborate network of expertise with the collaboration of researchers from various disciplines, including medicine, chemistry, bioengineering, and electronics may enable us to overcome future challenges. Acknowledgments The authors gratefully acknowledge use of the services and facilities of the Koç University Research Center for Translational Medicine (KUTTAM), funded by the Presidency of Turkey, Presidency of Strategy and Budget. The content is solely the responsibility of the authors and does not necessarily represent the official views of the Presidency of Strategy and Budget. The authors would also like to thank Ph.D. student Uğur Akcan for his contribution to the preparation of schematic illustrations of the manuscript. Disclosure of potential conflicts of interest No potential conflicts of interest were disclosed. References 1.Zlokovic BV. Neurovascular pathways to neurodegeneration in Alzheimer’s disease and other disorders. Nat Rev Neurosci. 2011;12(12):1–20. doi: 10.1038/nrn3114. [DOI] [PMC free article] [PubMed] [Google Scholar] 2.Pardridge WM. Blood–brain barrier endogenous transporters as therapeutic targets: a new model for small molecule CNS drug discovery. 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[DOI] [PubMed] [Google Scholar] Articles from Tissue Barriers are provided here courtesy of Taylor & Francis ACTIONS View on publisher site PDF (3.2 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page ABSTRACT Introduction Neurovascular unit Barrier type endothelial cells Basement membrane Pericytes Astrocytes Microglia Neuron Functions of the blood-brain barrier Characteristics of the blood-brain barrier Transport pathways across the blood-brain barrier The paracellular pathway The transendothelial pathway Cell movement across the BBB Circumventricular organs BBB disruption in pathological conditions Drug delivery into the brain for clinic implications Future challenges Acknowledgments Disclosure of potential conflicts of interest References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
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https://www.kenhub.com/en/library/anatomy/semispinalis-cervicis-muscle
Register now and grab your free ultimate anatomy study guide! Semispinalis cervicis muscle Author: Jana Vasković, MD • Reviewer: Francesca Salvador, MSc Last reviewed: November 03, 2023 Reading time: 3 minutes Semispinalis cervicis is a muscle of the back. Along with the semispinalis capitis and thoracis it comprises a long semispinalis muscle. This muscle belongs to the spinotransverse group of deep back muscles, along with the rotatores and multifidus muscles. The main anatomical feature of the semispinalis muscle, and of the semispinalis cervicis, is that it attaches between the spinous and transverse processes of vertebrae, bridging five to six vertebral levels along its course. Semispinalis cervicis acts in synergy with its capitis and thoracis counterparts. Their conjoint function facilitates the movements of the vertebral column; extension, lateral flexion and rotation of the head, cervical and thoracic spines. This article will discuss the anatomy and function of the semispinalis cervicis muscle. Key facts about the semispinalis cervicis muscle | | | --- | | Origin | Transverse processes of vertebrae T1-T6 | | Insertion | Spinous processes of vertebrae C2-C5 | | Action | Bilateral contraction - Extension of head, cervical and thoracic spine Unilateral contraction - Lateral flexion of head, cervical and thoracic spine (ipsilateral), rotation of head, cervical and thoracic spine (contralateral) | | Innervation | Medial branches of posterior rami of spinal nerves | | Blood supply | Occipital, deep cervical and vertebral arteries | Anatomy Semispinalis cervicis originates from the transverse processes of vertebrae T1-T6. The slender fascicles of this muscle span over the thoracic and cervical regions of the back to finally insert onto the posterior surfaces of spinous processes of vertebrae C2-C5. The muscle fibers span in a laminated fashion, so that those originating from T1 insert to the C2, while those stemming from T6 insert to the C5. The dorsal and lateral aspects of the semispinalis cervicis lie deep to the semispinalis capitis muscle. On its course, the muscle covers the dorsal surface of multifidus cervicis and thoracis muscles, while medially are the bodies of cervical and thoracic vertebrae. Semispinalis cervicis is innervated by the medial branches of posterior rami of spinal nerves, while its arterial supply comes from the occipital, deep cervical and vertebral arteries. Function The function of semispinalis cervicis is to help the head and neck extensors to extend the head at the neck when contracting bilaterally. On the other hand, its unilateral contraction aids the same muscles to perform the ipsilateral lateral flexion and contralateral rotation of the head, cervical and thoracic spines. Sources References: Layout: Illustrators: Semispinalis cervicis muscle: want to learn more about it? Our engaging videos, interactive quizzes, in-depth articles and HD atlas are here to get you top results faster. What do you prefer to learn with? “I would honestly say that Kenhub cut my study time in half.” – Read more. Kim Bengochea, Regis University, Denver Register now and grab your free ultimate anatomy study guide! Grounded on academic literature and research, validated by experts, and trusted by more than 6 million users. Read more. Kenhub fosters a safe learning environment through diverse model representation, inclusive terminology and open communication with our users. Read more. Follow us for daily anatomy content Learning anatomy isn't impossible. We're here to help. Learning anatomy is a massive undertaking, and we're here to help you pass with flying colours. Want access to this video? ...it takes less than 60 seconds! Want access to this quiz? ...it takes less than 60 seconds! Want access to this gallery? ...it takes less than 60 seconds!
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https://ocw.mit.edu/courses/18-125-measure-and-integration-fall-2003/resources/18125_lec14/
�� � � � � MEASURE AND INTEGRATION: LECTURE 14 Convex functions. Let ϕ : (a, b ) → R, where −∞ ≤ a < b ≤ ∞ . Then ϕ is convex if ϕ((1 − t)x + ty ) ≤ (1 − t)ϕ(x) + tϕ (y) for all x, y ∈ (a, b ) and t ∈ [0 , 1]. Looking at the graph of ϕ, this means that (t, ϕ (t)) lies below the line segment connecting (x, ϕ (x)) and (y, ϕ (y)) for x < t < y . Convexity is equivalent to the following. For a < s < t < u < b , ϕ(t) − ϕ(s) ϕ(u) − ϕ(t) . t − s ≤ u − t If ϕ is differentiable, then ϕ is convex on (a, b ) if and only if, for a < s < t < b , ϕ�(s) ≤ ϕ�(t). If ϕ is C2 (continuously twice differentiable), then ϕ� increasing ⇒ ϕ�� ≥ 0. Theorem 0.1. If ϕ is convex on (a, b ), then ϕ is continuous on (a, b ). Jensen’s inequality. Let (Ω , M, μ ) be a measure space such that μ(Ω) = 1 (i.e., μ is a probability measure). Let f : Ω → R and f ∈ L1(μ). If a < f (x) < b for all x ∈ Ω and ϕ is convex on (a, b ), then ϕ f dμ (ϕ ◦ f )dμ. ≤ ΩΩ Proof. Let t = Ω f dμ . Since a < f < b , a = a · μ(Ω) < f dμ < b · μ(Ω) = b, Ω so a < t < b . Conversely, ϕ(t) − ϕ(s) ϕ(u) − ϕ(t) . t − s ≤ u − t Fix t, and let ϕ(t) − ϕ(s) B = sup . a<s<t t − s Then ϕ(t) − ϕ(s) ≤ B(t − s) for s < t . We have ϕ(u) − ϕ(t) B ≤ u − t Date :October 21, 2003. 1 � � � �� � � � � � � � 2MEASURE AND INTEGRATION: LECTURE 14 for any u ∈ (t, b ), so B(u − t) ≤ ϕ(u) − ϕ(t) for u > t. Thus ϕ(s) ≥ ϕ(t) + B(s − t) for any a < s < b. Let s = f (x) for any x ∈ Ω. Then ϕ(f (x)) − ϕ(t) − B(f (x) − t) ≥ 0 for all x ∈ Ω. Now ϕ convex ϕ continuous, so ϕ f is measurable. Thus, inte ⇒ ◦ grating with respect to μ, (ϕ ◦ f )dμ − ϕ(t) dμ − B f dμ ≥ 0, XXX and the inequality follows. � Examples. (1) Let ϕ(x) = ex be a convex function. Then exp f dμ ef dμ. ≤ ΩΩ (2) Let Ω = {p1, . . . , p n} be a finite set of points and define μ({pi}) = 1/n . Then μ(Ω) = 1. Let f : Ω R with f (pi) = xi. Then → � n f dμ = f (pi)μ({pi}) Ωi=1 1 + xn).= n (x1 + · · · Thus 1 exp + xn) ≤ ef dμ n (x1 + · · · Ω 1 (ex1 + exn ).≤ n + · · · xi Let yi = e . Then 1 yn)1/n + yn),(y1 + · · · ≤ n (y1 + · · · which is the inequality between arithmetic and geometric means. We also could take μ({pi}) = αi > 0 and n αi = 1. Then i=1 α1α2αn y1 y2 · · · yn ≤ α1y1 + α2y2 + · · · + αnyn.� � � � � � � � � � MEASURE AND INTEGRATION: LECTURE 14 3 H¨ older’s and Minkowski’s inequalities. We define numbers p and q to be conjugate exponents if 1/p + 1/q = 1. The conjugate exponent of 1 is ∞. Conjugate exponents are the same if and only if p = q = 2. Theorem 0.2. Let p and q be conjugate exponents with 1 < p < ∞. Let (X, M, μ) be a measure space and f, g : X → [0 , ∞] measurable functions. Then � �� �1/p �� �1/q f g dμ ≤ f p dμ gq dμ (H¨ older’s) XXX and �� �1/p �� �1/p �� �1/p (f + g)pdμ ≤ X f p dμ + gp dμ (Minkowski’s) . XX Proof. H¨ � older’s. Without loss of generality we may assume that f p =� � X 1 and X gq = 1. Indeed, if f p = 0 and � gq = 0, then let f g f = �� �1/p , g = �� .�1/p f p f p XX (Otherwise, if f p = 0, then f p = 0 a.e., and both sides of the inequal ity are equal to zero.) We claim that 1 1 (0.1) ab ≤ ap + bq for all a, b ∈ [0 , ∞]. p q It is easy to check if a or b equals 0 or ∞. Assume 0 < a < ∞ and 0 < b < ∞, and write a = es/p and b = et/q for some s, t ∈ R. Let Ω = {x1, x2}, μ(x1) = 1/p , and μ(x2) = 1/q . We have exp f dμ ≤ ef dμ, ΩΩ where f (x1) = s and f (x2) = t. Thus, s t 1 1s exp + ≤ pe + et , p q q so (0.1) follows. Thus, 1 1 1 1 (f g)dμ ≤ ap dμ + bq dμ = + = 1. q X p qX p X Minkowski’s. Observe that (f + g)p = f (f + g)p−1 + g(f + g)p−1 .�� � � � � � � 4MEASURE AND INTEGRATION: LECTURE 14 Since p and q are conjugate exponents, q = p/ (p − 1). Thus, � �� �1/p �� �(p−1) /p (f + g)(p−1) p/ (p−1) f (f + g)p−1 ≤ f p �� �1/p �� �(p−1) /p = f p (f + g)p . Similarly, � �� �1/p �� �(p−1) /p (f + g)p .f (f + g)p−1 ≤ f p Let Ω = {x1, x2}, μ(x1) = 1/2 = μ(x2), and ϕ = tp. Then p f dμ f p dμ, ≤ ΩΩ so � � a + b p ap bp . 2 ≤ 2 2 Thus, 1 1 1 (f + g)p gp < ∞. 2p ≤ 2 f p + 2 Since 1 − (p − 1) /p = 1/p , �� �1/p �� �1/p �� �1/p (f + g)pdμ f p dμ + gp dμ .≤ XX X
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https://www.quora.com/What-is-the-angle-between-the-edge-and-the-plane-in-a-tetrahedron-and-how-do-you-get-to-it
Something went wrong. Wait a moment and try again. Solid Modelling Planes (geometry) 3D Geometry Analytic Solid Geometry Concept of Geometry Euclidean Geometry 5 What is the angle between the edge and the plane in a tetrahedron, and how do you get to it? James McMenemey Studied Physics & Education · Author has 113 answers and 188.8K answer views · 6y There is something missing here; I presume this refers to a regular tetrahedron. That makes it a very nice question in 3-dimensional geometry, and if you can handle that, or at least understand the theorem of Pythagoras, you can try the actual calculation yourself. One hint may help you on your way. Take any cube, and draw one single diagonal on each of its six faces in such a way as to join up four of its corners. Each of these diagonals meets two others at each end, and the other four corners of the cube are missed by these six diagonals. So now we have the six lines or edges, and four vertic There is something missing here; I presume this refers to a regular tetrahedron. That makes it a very nice question in 3-dimensional geometry, and if you can handle that, or at least understand the theorem of Pythagoras, you can try the actual calculation yourself. One hint may help you on your way. Take any cube, and draw one single diagonal on each of its six faces in such a way as to join up four of its corners. Each of these diagonals meets two others at each end, and the other four corners of the cube are missed by these six diagonals. So now we have the six lines or edges, and four vertices of a well known 3-D shape. Enjoy working it out! Related questions What is the internal angle between the two faces of a regular tetrahedron? The equilateral tetrahedron edge is 10 cm. What is the angle between this edge and the face that does not contain it? Does there exist a tetrahedron, so that every edge is the side of an obtuse angle of a face? A friend doesn't get that you can have a 70 degree angle inside a regular tetrahedron. She says the angle should be less than 60. How do I explain this? What is the angle between the edge BC and base OAB of the tetrahedron? Given that OA= I+3j-k,OB=3i-j+2k, OC=i-j-2k and OC is perpendicular to both OA and OB. How do you calculate the area of the base OAB and the volume of the tetrahedron? Peter Conrad I got 800 on my math SAT, back when that was the best score you could get. · Author has 682 answers and 1.2M answer views · 6y I like James M’s approach. Here is another possible starting point: Imagine a regular tetrahedron sitting on a flat surface. The vertex that is above the plane of the other three vertices will be directly above the centroid of the triangle formed by the edges that connect the three vertices on the plane. The distance from a vertex of this triangle to its centroid is one third of the triangle’s altitude. This should provide you with all the distances to find the angle between the edge and the plane. Buddha Buck Studied at University at Buffalo · Author has 5.8K answers and 16.9M answer views · 8y Related A friend doesn't get that you can have a 70 degree angle inside a regular tetrahedron. She says the angle should be less than 60. How do I explain this? I want to start by imagining a regular triangular prism instead of a regular tetrahedron. Two faces of the prism are regular triangles, and three faces are squares. It looks like this: Because of the construction, the dihedral angle between two of the yellow sides is clearly 60 degrees (the same as the internal angles on the equilateral triangle end-caps. Now imagine drawing on the top two yellow faces an equilateral triangle, using the top edge as a common side to the two triangles. It’s clear that the two lower corners of those equilateral triangles, plus the top two vertices of the prism, mak I want to start by imagining a regular triangular prism instead of a regular tetrahedron. Two faces of the prism are regular triangles, and three faces are squares. It looks like this: Because of the construction, the dihedral angle between two of the yellow sides is clearly 60 degrees (the same as the internal angles on the equilateral triangle end-caps. Now imagine drawing on the top two yellow faces an equilateral triangle, using the top edge as a common side to the two triangles. It’s clear that the two lower corners of those equilateral triangles, plus the top two vertices of the prism, make the corners of a tetrahedron — and that that tetrahedron has two faces with a 60 degree dihedral angle. But that tetrahedron is not regular: 5 of the edges are, by construction, equal in length, but not the sixth. It is shorter. In order to make it equal in length, the dihedral angle between the two original yellow faces needs to be increased — to about 70 degrees (Wikipedia says cos−1(13)≈70.53∘). Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix. Don’t waste your time Here’s the thing: I wish I had known these money secrets sooner. 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On two faces draw a perpendicular from the apex to the centre of the base of the triangle, ensuring that the two perpendiculars join at their respective bases. From Pythagoras theorem, the length of the perpendiculars is sqrt(3). A triangle is formed, therefore, by the two Inform your colleague thusly: Imagine the regular tetrahedron to have faces of equilateral triangles with sides of two units. On two faces draw a perpendicular from the apex to the centre of the base of the triangle, ensuring that the two perpendiculars join at their respective bases. From Pythagoras theorem, the length of the perpendiculars is sqrt(3). A triangle is formed, therefore, by the two perpendiculars and one sloping edge of the tetrahedron. The triangle has two sides of sqrt(3... Assistant Bot · 1y To find the angle between an edge and a plane in a tetrahedron, you can follow these steps: Understand the Geometry of the Tetrahedron: A tetrahedron consists of four triangular faces, six edges, and four vertices. The angle we want to find is between an edge and a plane formed by two of the other edges. Identify the Edge and the Plane: Suppose we have a tetrahedron with vertices A,B,C,D. If we want to find the angle between edge AB and the plane formed by triangle BCD, we consider the following: Edge AB is the line segment connecting points A and B. Plane BCD is defined by points B,C, and To find the angle between an edge and a plane in a tetrahedron, you can follow these steps: Understand the Geometry of the Tetrahedron: A tetrahedron consists of four triangular faces, six edges, and four vertices. The angle we want to find is between an edge and a plane formed by two of the other edges. Identify the Edge and the Plane: Suppose we have a tetrahedron with vertices A,B,C,D. If we want to find the angle between edge AB and the plane formed by triangle BCD, we consider the following: Edge AB is the line segment connecting points A and B. Plane BCD is defined by points B,C, and D. Find the Normal Vector of the Plane: To find the angle between the edge and the plane, we first need the normal vector of the plane BCD. This can be done by taking two vectors in the plane: →BC=C−B →BD=D−B The normal vector →N to the plane can be found using the cross product: →N=→BC×→BD Find the Direction Vector of the Edge: The direction vector of edge AB is: →AB=B−A Calculate the Angle: The angle θ between the edge AB and the plane BCD can be found using the dot product of the direction vector of the edge and the normal vector of the plane: cos(θ)=|→AB⋅→N||→AB||→N| To find θ: θ=arccos(|→AB⋅→N||→AB||→N|) Interpret the Result: The angle θ calculated will give you the angle between the edge AB and the normal to the plane. The angle between the edge and the plane itself is given by: ϕ=90∘−θ Conclusion By following these steps, you can calculate the angle between an edge and a plane in a tetrahedron. Remember that the specific coordinates of the vertices will determine the vectors used in your calculations. Related questions What is the sum of all the angles of the faces of a triangular pyramid? A triangular-based pyramid (or tetrahedron) has three right angles meeting at one corner. How is the area of the face opposite the right angles related to the areas of the other three faces? What is the angle between two plane mirrors when there are five images? In the tetrahedron, three triangles meet at each vertex. What is the sum of the interior angles? Is it less than 360 degrees? ABCD is a regular tetrahedron. If the side length AB = 6, what is the radius of the sphere inscribed into the tetrahedron? Philip Lloyd Former Specialist Calculus Teacher and Mentor.. · Author has 6.8K answers and 52.8M answer views · 5y Related A triangular-based pyramid (or tetrahedron) has three right angles meeting at one corner. How is the area of the face opposite the right angles related to the areas of the other three faces? I have drawn 2 views of the tetrahedron ABCD. Sides AB = BC = BD = x ———————————————————————————————— I have drawn 2 views of the tetrahedron ABCD. Sides AB = BC = BD = x ———————————————————————————————— Sponsored by CDW Corporation Want document workflows to be more productive? The new Acrobat Studio turns documents into dynamic workspaces. Adobe and CDW deliver AI for business. Gary Ward MaEd in Education & Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.6M answer views · 4y Related ABCD is a tetrahedron, where angle ABC = angle ABD = angle CBD = 90, AB=6cm and BC=BD=4cm. What is the angle between the planes ACD and BCD, correct to the nearest degree? ABCD is a tetrahedron, where angle ABC = angle ABD = angle CBD = 90, AB=6cm and BC=BD=4cm. What is the angle between the planes ACD and BCD, correct to the nearest degree? The distance from B to the mid-point of CD is 4sqrt(2)/2 = 2sqrt(2) tan-1(6/(2sqrt(2)) = 65 degrees to the nearest degree. ABCD is a tetrahedron, where angle ABC = angle ABD = angle CBD = 90, AB=6cm and BC=BD=4cm. What is the angle between the planes ACD and BCD, correct to the nearest degree? The distance from B to the mid-point of CD is 4sqrt(2)/2 = 2sqrt(2) tan-1(6/(2sqrt(2)) = 65 degrees to the nearest degree. Harish Chandra Rajpoot authored 'Advanced Geometry' on research articles in Mathematics & Radiometry · Author has 1.6K answers and 7.7M answer views · 4y Related How do I find out the solid angle subtended by a tetrahedron at its vertex? Case-1: If α,β & γ are the angles between consecutive lateral edges meeting at a vertex of a tetrahedron then the solid angle subtended by the tetrahedron at that vertex is given by following Generalized Formula ω=π2−sin−1(cosα−cosβcosγsinβsinγ)−sin−1(cosβ−cosαcosγsinαsinγ)−sin−1(cosγ−cosαcosβsinαsinβ) Case-2: If tetrahedron is a regular tetrahedron then substituting α=β=γ=π3 (i.e. the Case-1: If α,β & γ are the angles between consecutive lateral edges meeting at a vertex of a tetrahedron then the solid angle subtended by the tetrahedron at that vertex is given by following Generalized Formula ω=π2−sin−1(cosα−cosβcosγsinβsinγ)−sin−1(cosβ−cosαcosγsinαsinγ)−sin−1(cosγ−cosαcosβsinαsinβ) Case-2: If tetrahedron is a regular tetrahedron then substituting α=β=γ=π3 (i.e. the angles between consecutive lateral edges meeting at a vertex of a regular tetrahedron) then the solid angle subtended by a regular tetrahedron at any of its vertices is ω=π2−sin−1(cosπ3−cosπ3cosπ3sinπ3sinπ3)−sin−1(cosπ3−cosπ3cosπ3sinπ3sinπ3)−sin−1(cosπ3−cosπ3cosπ3sinπ3sinπ3) =π2−3sin−1(13) References: : Solid angles subtended by the platonic solids (regular polyhedrons) at their vertices by HCR : HCR's Theory of Polygon (solid angle subtended by any polygonal plane at any point in the space) Chad Hanna Volunteer Family Historian/genealogist and Teacher (1989–present) · Author has 2.8K answers and 1.3M answer views · 5y Related What degree angle are the corners of a regular tetrahedron? A regular tetrahedron is one of the 5 platonic solids and is made up of four equilateral triangles. For each equilateral triangle the edges are at 60 degrees to one another. However, a more interesting question is the angle between a triangle forming the base of the tetrahedron and one of the edges rising to the vertex. To calculate this we need two of the following: length of the triangle’s edge vertical height of the tetrahedron distance from the corner of the base triangle to the centre of the base triangle - which lies below the top of the tetrahedron The base of the triangle can split into thr A regular tetrahedron is one of the 5 platonic solids and is made up of four equilateral triangles. For each equilateral triangle the edges are at 60 degrees to one another. However, a more interesting question is the angle between a triangle forming the base of the tetrahedron and one of the edges rising to the vertex. To calculate this we need two of the following: length of the triangle’s edge vertical height of the tetrahedron distance from the corner of the base triangle to the centre of the base triangle - which lies below the top of the tetrahedron The base of the triangle can split into three triangles meeting at the centre of the base with angles 30, 30 and 120 degrees. Each of these can be split into two right triangles with one side being half the length of the edge. We assume the length of a triangle’s edge is 1, the distance to the centre of the triangle (the hypotenuse of the right triangles) can be calculated as 0.5/Cos 30 degrees. Cos 30 happens to be √3 divide by 2, so the distance is 0.5 x 2/√3 or 1/√3 The angle wanted is the arc cosine of adjacent/hypotenuse. The hypotenuse is an edge so it’s 1, the adjacent is 1/√3, so the cosine of the angle is 1/√3 = 0.5774 - and this corresponds to an angle of 54.7 degrees. 54.7 degrees for the angle between edge and base sounds about right because it must be a little less than the 60 degrees between the edges. Also, happily, this agrees with the Face-vertex-edge angle at Tetrahedron - Wikipedia Haresh Sagar Studied Science & Mathematics (Graduated 1988) · Author has 6.2K answers and 7M answer views · 2y Related Let M be the midpoint of line AB in regular tetrahedron ABCD. What is cos (angle CMD)? All the faces of a tetrahedron are equilateral triangles, therefore all the edges are equal. Since [math]M[/math] is midpoint of [math]AB[/math], [math]CM[/math] and [math]DM[/math] are heights so, [math]Cos(CMD)=\dfrac{\dfrac{3}{4}+\dfrac{3}{4}-1}{2\cdot\dfrac{3}{4}}=\dfrac{1}{3}\approx{70.53°}[/math] All the faces of a tetrahedron are equilateral triangles, therefore all the edges are equal. Since [math]M[/math] is midpoint of [math]AB[/math], [math]CM[/math] and [math]DM[/math] are heights so, [math]Cos(CMD)=\dfrac{\dfrac{3}{4}+\dfrac{3}{4}-1}{2\cdot\dfrac{3}{4}}=\dfrac{1}{3}\approx{70.53°}[/math] Rod Rishworth Studied at University of Cambridge · Author has 3.5K answers and 6.1M answer views · 7y Related Is it a coincidence that the dihedral angles of a regular tetrahedron and octahedron are supplementary? An interesting question, probably answered most easily by looking at Table of polyhedron dihedral angles - Wikipedia and noting that the angles between the planes for a couple of other polyhedra are the same, including the Tetratetrahedron (which I have never heard of before attempting to answer this question). Note that they all have triangular sides. Consider pasting tetrahedra to the sides of an octahedron. Use an octahedron with vertices at points on each of the (x, y, z) axes and distance 1 from the origin. Take as base for a tetrahedron the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). The f An interesting question, probably answered most easily by looking at Table of polyhedron dihedral angles - Wikipedia and noting that the angles between the planes for a couple of other polyhedra are the same, including the Tetratetrahedron (which I have never heard of before attempting to answer this question). Note that they all have triangular sides. Consider pasting tetrahedra to the sides of an octahedron. Use an octahedron with vertices at points on each of the (x, y, z) axes and distance 1 from the origin. Take as base for a tetrahedron the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). The final vertex is going to have equal x, y and z coordinates, all positive since we are pasting it on. Now imagine what is going to happen if we paste three more tetrahedrons of the same size, in those octants where two out of three coordinates are negative, that is to say (t, -t, -t), (-t, t, -t), and (-t, -t, t), where t is the value I didn’t bother to calculate in the previous paragraph for the point (t, t, t). Finally consider the distances between each of these pairs of points. They all share precisely one coordinate with each of the others, and the difference of the other two is each 2t. Again, it doesn’t matter what t is, because a shape which has four points all equidistant is a tetrahedron. So call it what you want, but an octahedron can be seen as a tetrahedron with its four “points” chopped off, or as the intersection of the two tetrahedra formed by taking alternate sets of vertices of a cube - hold a six-sided die by two opposite corners of the 1 side and then the opposite corners on the 2 and 5 sides of the ones you’ve picked to convince yourself that there are two such tetrahedrons. In summary, no, it isn’t a coincidence. Chris Manning B. Math. (Hons.) in Mathematics, The University of Newcastle (Australia) (Graduated 1981) · Author has 1.8K answers and 4.2M answer views · 5y Related What is the internal angle between the two faces of a regular tetrahedron? In x-y-z space, the figure whose vertices are O(0, 0, 0), A(1, 0, 0), B(1/2, (sqrt 3)/2, 0) and C(1/2, 1/(2 sqrt 3), sqrt(2/3)) is a regular tetrahedron. OAB is the base while C is the apex. The midpoint of OA is D(1/2, 0, 0). The point on the base directly under C is E(1/2, 1/(2 sqrt 3), 0). The internal angle between 2 faces = angle CDE = tan^(-1){sqrt(2/3)/[1/(2 sqrt 3)]} = tan^(-1)(2 sqrt 2) = approx. 1.231 radians or approx. 70.53 degrees. Juan E. Iglesias PhD in Chemical Engineering, The University of Texas at Austin (Graduated 1971) · Author has 493 answers and 133.6K answer views · 3y Related How do you find the angle between lines joining tetrahedron center to vertices (geometry, polyhedra, math)? Consider the cube whose vertices have the coordinates math[/math]. The vertices math[/math], math[/math], math[/math] and math[/math] define a regular tetrahedron having an edge of length [math]\sqrt{2}[/math]. The required angle can be computed by finding the inner product (dot product) of any pair of vectors with components equal to those of the vertices of the tetrahedron; the dot producto of the first two above mentioned is clearly [math]-1/4[/math], and their moduli is half the body diagonal of the cube, [math]\sqrt{3}/2[/math]. Hence [math]\cos\theta=\frac{-1/4}{3/4}=-1/3[/math], and [math]\theta=\arccos(-1/3)= 109º28'[/math] Consider the cube whose vertices have the coordinates math[/math]. The vertices math[/math], math[/math], math[/math] and math[/math] define a regular tetrahedron having an edge of length [math]\sqrt{2}[/math]. The required angle can be computed by finding the inner product (dot product) of any pair of vectors with components equal to those of the vertices of the tetrahedron; the dot producto of the first two above mentioned is clearly [math]-1/4[/math], and their moduli is half the body diagonal of the cube, [math]\sqrt{3}/2[/math]. Hence [math]\cos\theta=\frac{-1/4}{3/4}=-1/3[/math], and [math]\theta=\arccos(-1/3)= 109º28'16.39"[/math]. Related questions What is the internal angle between the two faces of a regular tetrahedron? The equilateral tetrahedron edge is 10 cm. What is the angle between this edge and the face that does not contain it? Does there exist a tetrahedron, so that every edge is the side of an obtuse angle of a face? A friend doesn't get that you can have a 70 degree angle inside a regular tetrahedron. She says the angle should be less than 60. How do I explain this? What is the angle between the edge BC and base OAB of the tetrahedron? Given that OA= I+3j-k,OB=3i-j+2k, OC=i-j-2k and OC is perpendicular to both OA and OB. How do you calculate the area of the base OAB and the volume of the tetrahedron? What is the sum of all the angles of the faces of a triangular pyramid? A triangular-based pyramid (or tetrahedron) has three right angles meeting at one corner. How is the area of the face opposite the right angles related to the areas of the other three faces? What is the angle between two plane mirrors when there are five images? In the tetrahedron, three triangles meet at each vertex. What is the sum of the interior angles? Is it less than 360 degrees? ABCD is a regular tetrahedron. If the side length AB = 6, what is the radius of the sphere inscribed into the tetrahedron? What is the internal angle between the two faces of a regular tetrahedron? How is it determined? How do I find the angle of the rays between two plane mirrors? Is it a coincidence that the dihedral angles of a regular tetrahedron and octahedron are supplementary? If five edges of a tetrahedron have the same length 4, what is the maximum volume of the tetrahedron? How is tetrahedron pronounced? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://www.youtube.com/watch?v=00jbG_cfGuQ
ATP & Respiration: Crash Course Biology #7 CrashCourse 16600000 subscribers 114327 likes Description 11184504 views Posted: 12 Mar 2012 In which Hank does some push-ups for science and describes the "economy" of cellular respiration and the various processes whereby our bodies create energy in the form of ATP. Special thanks go to Stafford Fitness (www.staffordfitness.net) for allowing us to shoot the gym scenes in their facilities. This video uses sounds from Freesound.org Table of Contents: 1) Cellular Respiration 01:00 2) Adenosine Triphosphate 01:29 3) Glycolysis 4:13 A) Pyruvate Molecules 5:00 B) Anaerobic Respiration/Fermentation 5:33 C) Aerobic Respiration 6:45 4) Krebs Cycle 7:06 A) Acetyl COA 7:38 B) Oxaloacetic Acid 8:21 C) Biolography: Hans Krebs 8:37 D) NAD/FAD 9:48 5) Electron Transport Chain 10:55 6) Check the Math 12:33 Crash Course is on Patreon! You can support us directly by signing up at Want to find Crash Course elsewhere on the internet? Facebook - Twitter - Instagram - CC Kids: Transcript: Oh, hello there! Uh, I'm at the gym. I don't know why you're here, but I'm going to do some push-ups, so you can join me on the floor if you want. Now I'm not doing this just to show off or anything; I'm actually doing this for science. Okay. grunt See what happened there? My arms moved. My shoulders moved. My back and stomach muscles moved. My heart pumped blood to all of those different places. It's pretty neat, huh? Well, it turns out that how we make and use energy is a lot like sports or other kinds of exercise. It can be hard work and a little bit complicated, but if you do it right it comes with some tremendous payoffs. But unlike hitting a ball with a stick, it's so marvelously complicated and awesome that we're still unraveling the mysteries of how it all works, and it all starts with a marvelous molecule that is one of your best friends: ATP. [Theme Music] Today, I'm talking about the energy and the process our cells and other animals' cells go through to provide themselves with power. Cellular respiration is how we derive energy from the food that we eat, specifically from glucose since most of what we eat ends up as glucose. Here's the chemical formula for one molecule of glucose. In order to turn this glucose into energy, we're going to need to add some oxygen: six molecules of it, to be exact. Through cellular respiration, we're going to turn that glucose and oxygen into six molecules of CO2, six molecules of water, and some energy that we can use for doing all of our push-ups. So that's all well and good, but here's the thing: we can't just use that energy to run a marathon or something. First, our bodies have to turn that energy into a really specific form of stored energy called ATP, or adenosine triphosphate. You've heard me talk about this before. People often refer to ATP as "the currency of biological energy." Think of it as an American dollar. It's what you need to do business in the U.S. You can't just walk into a Best Buy with a handful of Chinese yen or Indian rupees and expect to be able to buy anything with them even though they technically are money. Same goes with energy; in order to be able to use it, our cells need energy to be transferred into adenosine triphosphate to be able to grow, move, create electrical impulses in our nerves and brains, everything. A while back, for instance, we talked about how cells use ATP to transport some kinds of materials in and out of its membranes. To jog your memory about that, you can watch that episode right here. Now before we see how ATP is actually put together, let's look at how cells can cash in on the energy that's stashed in there. Well, adenosine triphosphate is made up of a nitrogenous base called adenine with a sugar called ribose and three phosphate groups attached to it. Now one thing you need to know about these three phosphate groups is that they are super uncomfortable sitting together in a row like that, like three kids on a bus who hate each other all sharing the same seat. So because the phosphate groups are such terrible company for each other, ATP is able to do this nifty trick where it shoots one of the phosphate groups off the end of the seat, creating ADP, or adenosine di-phosphate, because now there are just two kids sitting on the bus seat. And this reaction when the third jerk kid is kicked off the seat: energy is released. And since there are a lot of water molecules just floating around nearby, an OH pairing⁠—that's called a hydroxide⁠—from one of the H2Os comes over and takes the place of that third phosphate group, and everybody is much happier. By the way, when you use water to break down a compound like this, it's called hydrolysis, "hydro-" from water and "-lysis" from the Greek word for "separate." So now that you know how ATP is spent, let's see how it is minted, nice and new, by cellular respiration. Like I said, it all starts with oxygen and glucose. In fact, text books make a point of saying that through cellular respiration one molecule of glucose can yield a bit of heat and thirty-eight molecules of ATP. Now, it's worth noting that this number is kind of a best-case scenario; usually it's more like twenty-nine or thirty ATPs, but whatever! People are still studying this stuff, so let's stick with that number, thirty-eight. Now, cellular respiration isn't something that just happens all at once. Glucose is transformed into ATPs over three separate stages: glycolysis, the Krebs Cycle, and the electron transport chain. Traditionally, these stages are describes as coming one after the other, but really everything in the cell is kind of happening all at the same time. But let's start with the "first" step: glycolysis, or the breaking down of the glucose. Glucose, of course, is a sugar; you know this because it's got an "-ose" at the end of it. And glycolysis is just the breaking up of glucose's six-carbon ring into two three-carbon molecules called pyruvic acids, or pyruvate molecules. Now in order to explain how exactly glycolysis works, I'd need about an hour of your time, and a giant cast of finger puppets each playing a different enzyme, and though it would pain me to do it, I'd have to use words like phosphoglucoisomerase. But one simple way of explaining it is this: If you wanna make money, you gotta spend money. Glycolysis needs the investment of 2 ATPs in order to work, and in the end it generates 4 ATPs, for a net profit, if you will, of 2 ATPs. In addition to those 4 ATPs, glycolysis also results in 2 pyruvates and 2 super-energy-rich morsels called NADH, which are sort of the love-children of a B vitamin called NAD+ pairing with energized electrons and a hydrogen to create storehouses of energy that will later be tapped to make ATP. To help us keep track of all of the awesome stuff we're making here, let's keep score. So far we've created 2 molecules of ATP and 2 molecules of NADH, which will be used to power more ATP production later. Now, a word about oxygen. Like I mentioned, oxygen is necessary for the overall process of cellular respiration. But not every stage of it. Glycolysis, for example, can take place without oxygen, which makes it an anaerobic process. In the absence of oxygen, the pyruvates formed through glycolysis get rerouted into a process called fermentation. If there's no oxygen in the cell, it needs more of that NAD+ to keep the glycolysis going. So fermentation frees up some NAD+, which happens to create some interesting by products. For instance, in some organisms, like yeasts, the product of fermentation is ethyl alcohol, which is the same thing as all of this lovely stuff. But luckily for our day-to-day productivity, our muscles don't make alcohol when they don't get enough oxygen. If that were the case, working out would make us drunk, which actually would be pretty awesome, but instead of ethyl alcohol, they make lactic acid. Which is what makes you feel sore after that workout that kicked your butt. So, your muscles used up all the oxygen they had, and they had to kick into anaerobic respiration in order to get the energy that they needed, and so you have all this lactic acid building up in your muscle tissues. Uhhh! Uhhhhh! Uhhhhhh! Back to the score. Now we've made 2 molecules of ATP through glycolysis, but your cells really need the oxygen in order to make the other 30-some molecules they need. And that's because the next two stages of cellular respiration⁠—the Krebs Cycle and the electron transport chain⁠—are both aerobic processes, which means they require oxygen. And so we find ourselves at the next step in cellular respiration: after glycolysis comes the Krebs Cycle. So, while glycolysis occurs in the cytoplasm, or the fluid medium within the cell that all the organelles hang out in, the Krebs Cycle happens across the inner membrane of the mitochondria, which are generally considered the power centers of the cell. The Krebs Cycle takes the products of glycolysis⁠—those carbon-rich pyruvates⁠—and reworks them to create another 2 ATPs per glucose molecule, plus some energy in a couple of other forms, which I'll talk about in a minute. Here's how: First, one of the pyruvates is oxidized, which basically means it's combined with oxygen. One of the carbons off the three-carbon chain bonds with an oxygen molecule and leaves the cell as CO2. What's left is a two-carbon compound called acetyl coenzyme A, or acetyl coA. Then, another NAD+ comes along, picks up a hydrogen and becomes NADH. So our two pyruvates create another 2 molecules of NADH to be used later. As in glycolysis, and really all life, enzymes are essential here. They are the proteins that bring together the stuff that needs to react with each other, and they bring them together in just the right way. These enzymes, for example, bring together a phosphate with an ADP, to create another ATP molecule for each pyruvate. Enzymes also help join the acetyl coA and a 4-carbon molecule called oxaloacetic acid. I think that's how you pronounce it. Together they form a 6-carbon molecule called citric acid, and I'm certain that's how you pronounce that one because that's the stuff that's in orange juice. [Biolo-graphy Music] Fun fact: The Krebs Cycle is also known as the Citric Acid Cycle because of this very byproduct. However, it's usually referred to by the name of the man who figured it all out: Hans Krebs, an ear, nose, and throat surgeon who fled Nazi Germany to teach biochemistry at Cambridge, where he discovered this incredibly complex cycle in 1937. For being such a total freaking genius, he was awarded the Nobel Prize in Medicine in 1953. Anyway, the citric acid is then oxidized over a bunch of intricate steps, cutting carbons off left and right, to eventually get back to oxaloacetic acid, which is what makes the Krebs Cycle a cycle. And as the carbons get cleaved off the citric acid, there are leftovers in the form of CO2 or carbon dioxide, which are exhaled by the cell, and eventually by you. You and I, as we continue out existence as people, are exhaling the products of the Krebs Cycle right now. Good Work. Breathes This video, by the way, I'm using a lot of ATP making it. Now, each time a carbon comes off the citric acid, some energy is made, but it's not ATP. It's stored in a whole different kind of molecular package. This is where we go back to NAD+ and its sort of colleague FAD. NAD+ and FAD are chummy little enzymes that are related to B vitamins, derivatives of Niacin and Riboflavin, which you might have seen in the vitamin aisle. These B vitamins are good at holding on to high energy electrons and keeping that energy until it can get released later in the electron transport chain. In fact, they're so good at it that they show up in a lot of those high energy-vitamin powders that the kids are taking these days. NAD+s and FADs are like batteries, big awkward batteries that pick up hydrogen and energized electrons from each pyruvate, which in effect charges them up. The addition of hydrogen turns them into NADH and FADH2, respectively. Each pyruvate yields 3 NADHs and 1 FADH2 per cycle, and since each glucose has been broken down into two pyruvates, that means each glucose molecule can produce 6 NADHs and 2 FADH2s. The main purpose of the Krebs Cycle is to make these powerhouses for the next and final step, the Electron Transport Chain. And now comes the time when your saying, "Sweet pyruvate sandwiches, Hank, aren't we supposed to be making ATP here? Let's make it happen, Capt'n! What's the holdup?" Well friends, your patience is finally paying off, because when it comes to ATPs, the electron transport chain is the real moneymaker. In a very efficient cell, it can net a whopping 34 ATPs. So, remember all those NADHs and FADH2s that we made in the Krebs Cycle? Well, their electrons are going to provide the energy that will work as a pump along a chain of channel proteins across the inner membrane of the mitochondria where the Krebs Cycle occurred. These proteins will swap these electrons to send hydrogen protons from inside the very center of the mitochondria, across its inner membrane to the outer compartment of the mitochondria. But once they're out, the protons want to get back to the other side of the inner membrane, because there's a lot of other protons out there, and as we've learned, nature always tends to seek a nice, peaceful balance on either side of a membrane. So all of these anxious protons are allowed back in through a special protein called ATP synthase. And the energy of this proton flow drives this crazy spinning mechanism that squeezes some ADP and some phosphates together to form ATP. So, the electrons from the 10 NADHs that come out of the Krebs Cycle have just enough energy to produce roughly 3 ATPs each. And we can't forget our friends the FADH2s. We have two of them and they make 2 ATPs each. And voila! That is how animal cells the world over make ATP through cellular respiration. Now just to check, let's reset our ATP counter and do the math for a single glucose molecule once again: We made 2 ATPs for each pyruvate during glycolysis. We made 2 during the Krebs Cycle. And then during the electron transport chain we made about 34. And that's just for one molecule of glucose. Imagine how much your body makes and uses every single day. Don't spend it all in one place now! You can go back and watch any parts of this episode that you didn't quite get and I really want to do this quickly because I'm getting very tired. Pants If you want to ask us questions you can see us in the YouTube comments below and of course, you can connect with us on Facebook or Twitter. Grunts
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https://www.americanbookwarehouse.com/127011/?srsltid=AfmBOoptFMtXqFdfCKIzq1LQlGOjPmM73X-y-soZ2ZO2U0j9IaPU4Shn
Elements of Electromagnetics by Matthew N O Sadiku - American Book Warehouse Book Subjects Accounting Business Economics Education Engineering English Finance Government History Law Math Music Other Philosophy Reference Religion Science Social Science Technology Additional Information Additional Information Home About Us Contact Us Shipping & Returns FAQs Customer Reviews Book Deals Account Navigation Account Navigation My Account Sign in Create an account Call 1-615-200-6227 Currency - All prices are in AUD Currency - All prices in USD. Change Currency Loading... Please wait... Call us on 1-615-200-6227 My Account Sign in or Create an account About Us Contact Us Shipping & Returns FAQs Customer Reviews Book Deals Search Search Books Search Search Books 0 Book Subjects Accounting Business Economics Education Engineering English Finance Government History Law Math Music Other Philosophy Reference Religion Science Social Science Technology Elements of Electromagnetics by Matthew N O Sadiku ~~$129.00~~$29.95 (You save $99.05) Click to enlarge Elements of Electromagnetics by Matthew N O Sadiku ~~$129.00~~$29.95 (You save $99.05) SKU: Lowest Prices Online Over a Decade Selling Books Easy 30 Day Returns Huge Inventory to Search Fast Shipping Across U.S. Condition: USED - VERY GOOD Vendor: Weight: Availability: Shipping: Minimum Purchase: unit(s) Maximum Purchase: unit(s) : Gift Wrapping: Quantity: Buy in bulk and save Share Add to Wish List Click the button below to add the Elements of Electromagnetics by Matthew N O Sadiku to your wish list. Product Description Elements of Electromagnetics by Matthew N O Sadiku is available now for quick shipment to any U.S. location. This edition can easily be substituted for ISBN 0190698616 or ISBN 9780190698614 the 7th edition or even more recent edition. You will save lots of cash by using this 4th edition which is nearly identical to the newest editions. We have been selling books online for over ten years and we have learned how to save students from the inflated costs of textbooks especially when the updated editions do not contain substantial changes and typically are nearly identical in every way. We guarantee this by offering a 30-day full refund if you are unable to use the book for any reason. But we can assure you that this is a great way to save money and that this edition will work for you. If you need more convincing about our longstanding track record in saving students loads of unnecessary expense on books feel free to simply review nearly forty thousand positive reviews that can be seen on our Ebay store by clicking HERE Product Reviews Write Review Write Your Own Review How do you rate this product?Write a headline for your review here:Write your review here:Your email: We promise to never spam you, and just use your email address to identify you as a valid customer. Enter your name: (optional)Enter the code below: This product hasn't received any reviews yet. Be the first to review this product! Find Similar Products by Category Other Related Products Elements Of Electromagnetics by Matthew No Sadiku~~$158.95~~ $39.95 Elements Of Electromagnetics - Matthew N O Sadiku~~$121.50~~ $22.95 Elements Of Electromagnetics (Matthew Sadiku)~~$175.95~~ $36.95 Elements Of Electromagnetics by Matthew Sadiku~~$89.90~~ $35.95 Elements of Electromagnetics by Matthew Sadiku~~$239.99~~ $151.95 Company Info About Us Contact Us Shipping & Returns FAQs Customer Reviews Book Deals Newsletter Signup Name Email Secure Payments Connect With Us Facebook Twitter YouTube All prices are in USD.© 2025 American Book Warehouse|Sitemap|
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https://www.medicaljournals.se/jrm/content_files/download.php?doi=10.1080/16501970510043729
CASE REPORT MYASTHENIA GRAVIS MASQUERADING AS POST-POLIOMYELITIS SYNDROME Rajiv Singh and Brian Pentland From the Rehabilitation Medicine Unit, Astley Ainslie Hospital, Edinburgh, UK A 79-year-old man with previous bulbar poliomyelitis devel-oped dysphagia and was diagnosed as having post-polio syndrome. Over 2 years, his swallowing deteriorated and he suffered an aspiration pneumonia. Only after the subsequent development of fatigue and facial weakness was myasthenia gravis diagnosed. Diagnostic criteria for post-polio syndrome include the exclusion of all other neurological conditions such as myasthenia gravis. Moreover, in any instance where a patient develops new symptoms, it is advisable to reconsider the original diagnosis. Key words: myasthenia gravis, post-polio syndrome, dysphagia. J Rehabil Med 2006; 38: 136 / 137 Correspondence address: Raji v Singh, Rehabilitation Medicine Unit, Astley Ainslie Hospital, 133 Grange Road, Edinburgh EH9 2HL, UK. E-mail: raji v.singh@lpct.scot. nhs.uk Submitted January 20, 2005; accepted May 23, 2005 INTRODUCTION Muscular weakness arising many years, often decades, after recovery from acute poliomyelitis is sometimes referred to as post-polio syndrome (PPS) (1). While limb weakness is often prominent, dysphagia is described in up to half of those affected (2). Other features include fatigue, muscle or joint pain, sleep apnoea and respiratory insufficiency (3). In the USA, there are some 300,000 survivors of polio and up to 50% have features of PPS (4). There is therefore a high potential for attributing signs and symptoms to PPS that may arise from other pathology. We describe here a patient in whom the primary diagnosis of PPS masked the presence of myasthenia gravis. CASE REPORT A 79-year-old retired clergyman, who had suffered acute poliomyelitis with bulbar and limb involvement in 1947, presented to physicians with a 6-month history of dysphagia and reduced voice volume. A diagnosis of PPS was made and he was referred for speech and language therapy. Videofluoroscopy indicated poor pharyngeal contraction and clearance and he was given advice about swallowing strategies and speech intelligibility. His poor speech had been of particular concern given his reputation as a media personality. Little improvement occurred and 14 months later he developed a severe aspiration pneumonia, which required intravenous antibiotics. He was felt to present a high risk of aspiration and a percutaneous endoscopic gastrostomy tube was inserted. One year later, he developed fatigue and was seen in our neurorehabilitation clinic. On examination there was marked dysarthria, soft palate wasting, facial and jaw weakness but no demonstrable fatig-ability. There was no ptosis or diplopia and limb motor power was normal. At this stage, the alternative diagnosis of myasthenia gravis was considered. An assay for anti-acetylcholine receptor anti-bodies was strongly positive at 311 /10 10 mmol/l (normal 0 / 5) and an edrophonium test was positive. Electromyography showed significant reduction in muscle contraction on repetitive stimulation and increased ‘‘jitter’’ further supporting the diag-nosis. A computerized tomography scan of the thorax showed no evidence of thymoma. Within a day of starting pyridostigmine he noticed increased muscle strength and he continues to improve on a combination of steroids and pyridostigmine. DISCUSSION The presenting features of dysphagia and reduced speech volume in this patient were consistent with a diagnosis of PPS as was the later development of fatigue. There were no features such as ptosis, diplopia or limb weakness to suggest myasthenia gravis. Dysphagia and dysarthria are, however, recognized presenting features of this disease (5). The classical, though not universal characteristic of myasthenic muscle weakness is that motor activity fatigues. Thus the difficulty with eating and swallowing may be absent at the beginning of a meal but becomes evident as it progresses. This pattern was not present. The fatigue described by the patient was a general lethargy and malaise rather than a progressive weakness with activity. The development of facial weakness was the trigger to investigation for myasthenia gravis. High anti-acetylcholine receptor anti-body titres are common in young female patients or those with thymoma but has also been reported in elderly patients as in this case (6). A search of the principal medical databases (Medline, Embase, CINAHL) has failed to find a single report of the J Rehabil Med 2006; 38: 136 / 137 2006 Taylor & Francis. ISSN 1650-1977 DOI: 10.1080/16501970510043729 J Rehabil Med 38 co-existence of PPS and myasthenia gravis. We would recom-mend that clinicians retain a high index of suspicion as to the possibility of myasthenia gravis in patients with dysphagia occurring many years after acute poliomyelitis. This seems particularly pertinent as there are sensitive diagnostic tests available for myasthenia gravis, whereas PPS remains a clinical diagnosis lacking strict criteria. Furthermore, treatment is simple and can have a marked effect on strength and fatigue. In this instance early diagnosis and treatment may have prevented a life-threatening aspiration pneumonia. As a prin-ciple, it is worthwhile reconsidering an existing diagnosis when new features develop and to exclude the possibility of a new condition. REFERENCES 1. Trojan DA, Cashman NR. Pathophysiology and diagnosis of post-polio syndrome. Neurorehabilitation 1997; 8: 83 / 92. 2. Sonies BC, Dalakas MC. Dysphagia in patients with the post-polio syndrome. N Engl J Med 1991; 324: 1162 /1167. 3. Halstead LS, Grimby G. Post-polio syndrome. Philadelphia: Hanley & Belfus/Mosby; 1995, p. ix. 4. Silbergleit AK, Waring WP, Sullivan MJ, Maynard FM. Evaluation, treatment, and follow-up results of post polio patients with dyspha-gia. Otolaryngol Head Neck Surg 1991; 104: 333 / 338. 5. Colton-Hudson A, Koopman WJ, Moosa T, Smith D, Bach D, Nicolle M. A prospective assessment of the characteristics of dysphagia in myasthenia gravis. Dysphagia 2002; 17: 147 /151. 6. Vincent A, Palace J, Hilton-Jones D. Myasthenia gravis. Lancet 2001; 357: 2122 /2128. Myasthenia gra vis masquerading as post-poliomyelitis syndrome 137 J Rehabil Med 38
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https://www.youtube.com/watch?v=tXuyPiYvGZ4
Rational Exponents Nerdstudy 100000 subscribers 3953 likes Description 213774 views Posted: 26 Jul 2016 Watch this video to learn Rational Exponents to a deeper level of understanding! Nerdstudy aims to create the most appealing and informative educational video resources for students ranging from grade 7 to post secondary education. Subscribe to Nerdstudy's channel: Visit nerdstudy.com for more high quality lessons. Nerdstudy, the future of learning. 99 comments Transcript: today we're going to be learning about rational exponents so far the exponents that we've been considering were integers negative numbers zeroes positive numbers but we never looked at fractions as exponents for example if we have something like 4 to the exponent 1 over 2 what would that be it's certainly difficult to understand what this would mean but there is another equivalent form that might help us to understand what this means now this over here is equal to this so notice how this exponent over here has a numerator of 1 and a denominator of 2 what we end up with is square root of 4 2 the exponent 1 so once again you can see that from the exponent the denominator is what determined the degree of the root of course since the denominator is 2 we have a root of degree 2 more commonly known as a square root and since the numerator is 1 we are looking for the square root of 4 2 the exponent 1 and of course the square root of 4 can be further simplified to 2 great so if we had something like 16 to the exponent 1 over 2 then this would be the square root of 16 the exponent 1 of course that simplifies down to 4 and we can say this in a general sense notice here what is happening to be and what is happening to a now with that in mind let's try another example where we specifically try to use this general formula and you'll notice that it's actually really easy to do so if we had something like 32 to the exponent 3 over 2 that would just be the square root of 32 to the exponent 3 if it helps you to understand exponents better another way to do this would have been to see this 3 over 2 as the same thing as 3 times 1 over 2 after all 3 is equal to 3 over 1 and when you multiply the numerators and the denominators you would end up at 3 over 2 now if you decide to see the exponent as 3 times 1 over 2 then you might also be able to see that this is essentially a power of a power situation where you have 32 to the exponent 3 bracket it to the exponent 1 over 2 and now of course you see the similarities between this example and our very first example where we did 4 to the exponent 1 over 2 remember that when we raised 4 to the exponent of 1 over 2 it was the same thing as simply taking the square root of 4 similarly if we're raising 32 to the exponent 3 to the exponent of 1 over 2 then that would be the same as taking the square root of this entire power right here awesome so let's just do one last question let's make this a little bit more difficult but don't let that fool you this isn't actually as difficult as it might seem so we have something like this we see some fractions in the exponents well first all we have to do is deal with the operators negative 12 over 7 plus 2 over 7 well that's pretty simple that's just going to be negative 10 over 7 now the question is what do we do from here well we have already learned what to do with negative exponents 18 to the exponent negative 10 over 7 is going to be the exact same thing as 1 over 18 to the exponent 10 over Sonne remember we can just do one over and that entire power dropping that negative sign great now we're almost finished but we did learn something today we can put that 18 to the exponent 10 over 7 in a different form do we need to maybe not but if our teacher wants us to then maybe we should so in this situation we have 1 over 7th root of 18 to the exponent 10 now reversely if we saw something in its radical form we should be able to find out its equivalent exponential form as well so if we had something like the 4th root of 5 to the exponent 7 then of course this can be translated to 5 to the exponent 7 over 4 so it becomes pretty obvious then that this formula over here can help us out quite a bit just remember that they are two different ways to write the same thing the same value so there you have it now you know how to interpret powers that have fractions as their exponents and we will see you in our next video
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https://www.sciencedirect.com/topics/neuroscience/genetic-drift
Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Chapter Genetic Drift 2013, Brenner's Encyclopedia of Genetics (Second Edition)O. Honnay Abstract Next to mutation, gene flow, and natural selection, genetic drift is one of the four factors causing a gene pool to change over time. Genetic drift is defined as the random variation in allele frequencies between generations in finite populations, due to sampling error. Genetic drift is a nondirectional process, causing (1) loss of genetic variation from populations, (2) genetic differentiation among populations, and (3) increased homozygosity of the individuals. Even currently, large populations may have been subjected to genetic drift through bottleneck or founder effects. Genetic drift plays an important role in conservation biology where it is one of the factors that determines the minimal viable population size of a species. Genetic drift is also at the heart of the shifting-balance theory of evolution and of the neutral theory of molecular evolution. View chapterExplore book Read full chapter URL: Reference work2013, Brenner's Encyclopedia of Genetics (Second Edition)O. Honnay Chapter Volume 2 2019, Encyclopedia of Animal Behavior (Second Edition)Charles J. Goodnight, Gil Rosenthal Phase 1: The Phase of Genetic Drift During this first phase, evolution in small populations is dominated by genetic drift (Fig. 2(a)). Genetic drift is a function of population size: in very small populations, even selected alleles tend to behave as if they are neutral. Wright envisioned that this drift occurring in subpopulations allows each to move across the adaptive landscape independently. Indeed, some subpopulations potentially drift ‘down hill’ on the fitness slope and eventually cross an adaptive valley. Thus, genetic drift is the feature of the SBT allowing a subpopulation to escape the influence of one adaptive peak, move through an adaptive valley, and come under the selective influence of a new adaptive peak. View chapterExplore book Read full chapter URL: Reference work2019, Encyclopedia of Animal Behavior (Second Edition)Charles J. Goodnight, Gil Rosenthal Review article How maladaptation can structure biodiversity: eco-evolutionary island biogeography 2015, Trends in Ecology & EvolutionTimothy E. Farkas, ... Andrew P. Beckerman Genetic drift Genetic drift is a stochastic process that can generate maladaptation by causing the spread and fixation of deleterious alleles. Although drift is expected to be strongest for neutral loci, it can also influence loci under selection, potentially compromising local adaptation. Similarly, founder effects can be expected to generate (mal)adaptation in newly colonized habitats. In addition, genetic drift and/or founder effects coupled with inbreeding can generate inbreeding depression, which can be manifested as generalized maladaptation unrelated to any particular ecological environment . Genetic drift is most pronounced in small and isolated populations, and where gene flow from divergent populations is less likely (see also the next section). Thus, maladaptation arising from genetic drift will be greater, or more likely, in small and isolated populations than in large and well-connected populations. View article Read full article URL: Journal2015, Trends in Ecology & EvolutionTimothy E. Farkas, ... Andrew P. Beckerman Chapter Fundamentals of Molecular Evolution 2014, Bioinformatics for BeginnersSupratim Choudhuri 2.4.4 Genetic Drift Genetic drift (also called random genetic drift) means a change in the gene pool strictly by chance fixation of alleles. The effects of genetic drift can be acute in small populations and for infrequently occurring alleles, which can suddenly increase in frequency in the population or be totally wiped out. The alleles thus fixed by chance (genetic sampling error) may be neutral—that is, they may not confer any survival or reproductive advantage. Therefore, for small populations, genetic drift can result in a significant change in gene frequency in a short period of time. Genetic drift can be caused by a number of chance phenomena, such as differential number of offspring left by different members of a population so that certain genes increase or decrease in number over generations independent of selection, sudden immigration or emigration of individuals in a population changing gene frequency in the resulting population, or population bottleneck. Of these, population bottleneck can cause a radical change in allele frequencies in a very short time. A population bottleneck occurs when a population suddenly shrinks in size owing to random events, such as sudden death of individuals due to environmental catastrophe, habitat destruction, predation, or hunting. When the small number of surviving individuals gives rise to a new population, there is a radical change in the gene frequency in the resulting population, in which certain genes (including rare alleles) of the original population may radically increase in proportion while others may radically decrease or be wiped out completely, independently of selection. Additionally, the resulting population contains a small fraction of the genetic diversity of the original population. The founder effect is a severe case of population bottleneck and happens when a few individuals migrate out of a population to establish a new subpopulation. Random genetic drift accompanies such founder effect, to severely reduce the genetic variation that exists in the original population. In the new population, the founder effect can rapidly increase the frequency of an allele whose frequency was very low in the original population. If the allele is a disease-related allele, the founder effect can lead to the prevalence of the disease in the new population. An increase in a specific disease in a human population due to the founder effect is seen in the Old Order Amish of eastern Pennsylvania,66 and in the Afrikaner population of South Africa.67 The current Amish population has descended from a small number of German immigrants who settled in the United States during the eighteenth century. The incidence of Ellis–van Creveld syndrome (a form of dwarfism with polydactyly, abnormalities of the nails and teeth, and heart problems) is many times more prevalent in this Amish population than in the American population in general. The origin of this disease can be traced back to one couple, Samuel King and his wife, who came to the area in 1744. The mutated gene that causes the syndrome was passed along from the Kings and their offspring. The Amish population practices endogamy (individuals tend to mate within their own subgroup). Additionally, in this community the gene flow is centrifugal—that is, members may leave the community but outsiders do not join the community—therefore, there has been no introduction of exogenous genes into the Amish gene pool. As a result, the frequency of the disease gene has rapidly increased over generations. Another example of founder effect comes from the Afrikaner population of South Africa, which is mainly descended from one group of European (mainly Dutch, but also German and French) immigrants that landed there in 1652. The present-day Afrikaner population has a very high prevalence of Huntington’s disease; over 200 affected individuals in more than 50 supposedly unrelated families have been found to be ancestrally related through a common progenitor in the seventeenth century. Thus, the root of the disease can be traced back over 14 generations to a common progenitor who supposedly carried the gene for Huntington’s disease. Huntington’s disease is an autosomal dominant disease caused by triplet (CAG) repeat expansion in the gene (and the mRNA), containing 40 to>100 CAG triplets. The onset and severity of the disease is directly correlated with the number of repeats. View chapterExplore book Read full chapter URL: Book2014, Bioinformatics for BeginnersSupratim Choudhuri Chapter Genetics – Variation, Sexuality, and Evolution 2016, The Fungi (Third Edition)Lynne Boddy Genetic Drift and the Founder Effect Genetic drift is the change in frequencies of alleles in a population due to chance. If a population is small then chance could determine whether a neutral allele becomes extinct or increases in frequency to fixation. If a population is very small then such random genetic drift could determine the fate of an allele even in the presence of moderately strong natural selection. In nature, however, it may be unusual for a population to stay small long enough for drift to occur – the population could become extinct, grow, or merge with another population. Tendencies to genetic drift will be opposed by gene flow. Hence if a fungus is abundant and widespread with copious spores capable of long distance dispersal, gene flow is likely to counteract any tendency to genetic drift. There is evidence for this in the cosmopolitan and abundant fungi Neurospora crassa, Puccinia graminis f. sp. tritici, and Schizophyllum commune. There are, however, ways in which random events could determine the genetic structure of a population and the course of microevolution. One or a few individuals will not cover the genetic diversity in the population; many alleles present in the whole population will be absent from such a small set of individuals. A small set of individuals could occur as the result of a catastrophe almost destroying a population or by the dispersal of one or a few individuals to a new environment. The population resulting from such a founder effect will be genetically different from the one from which it originated. Many fungi live in environments that are highly favourable but transient, and will hence be liable to colonisation from one or a few spores when they arise, and population crashes when they disappear. Founder effects are likely to occur with such fungi and, if the fungi are not highly abundant, may not subsequently be overwhelmed by gene flow. Australia provides lots of examples of single founder events: Puccinia striiformis – cause of yellow (stripe) rust of wheat – was introduced into Australia in 1979 (p. 263), as a single race from Europe, but mutations have now resulted in new pathotypes which differ from those in Europe. Similarly, Cryphonectria parasitica – cause of chestnut blight (pp. 287–289) – in North America has much less genetic diversity than in Asia, probably reflecting a founder effect. The dry-rot fungus, Serpula lacrymans, originated in northeast Asia, where it has most genetic variations. However, there is very little genetic variation in the founder populations across the globe (Figure 4.11). In some areas the indoor genetic populations are unique (e.g. Japan), representing a single founder event, whereas elsewhere (e.g. Australia), there is slightly more variation representing founder events from Japan and from Europe. View chapterExplore book Read full chapter URL: Book2016, The Fungi (Third Edition)Lynne Boddy Chapter Population and Evolutionary Genetics 2012, Human Genes and GenomesLeon E. Rosenberg, Diane Drobnis Rosenberg Genetic Drift Chance or other unpredictable events can produce large changes in allele frequencies over a single generation. This is called genetic drift, and it usually has a much more profound effect on small populations than large ones. Drift often occurs because a newly formed, isolated group is not representative of the original population. Events like natural catastrophes (hurricanes) or epidemics may be responsible for creating such a genetic isolate. For example, the Pingelapese people occupy a group of Micronesian islands in the eastern Caroline Islands. This group has a frequency of 1 : 10 for an autosomal recessive form of blindness that occurs in other parts of that region with a frequency of 1 : 20,000 to 1 : 50,000. This situation resulted from a typhoon in 1780 that killed all but 9 males and 10 females among the Pingelapese. This catastrophe created what is referred to as a founder effect and a population bottleneck. A founder effect may lead to an unusually high frequency of a genetic disorder in a small isolated population. The catastrophe just described also created a “population bottleneck,” in which a population’s sample of alleles is changed by chance when the population is reduced in size suddenly. View chapterExplore book Read full chapter URL: Book2012, Human Genes and GenomesLeon E. Rosenberg, Diane Drobnis Rosenberg Chapter Neutral Mutation 2001, Brenner's Encyclopedia of Genetics (Second Edition)M. Kreitman Dynamics within Populations Natural selection cannot act to increase or decrease the frequency of a neutral mutation in a population by definition. Instead, the dynamics of allele frequency change of a neutral mutation is governed entirely by a process known as ‘genetic drift’. Genetic drift can be defined as the chance change in the frequency of a mutation in a population from one generation to the next resulting from the finite size of a population. Genetic drift will be strongest in populations of small size and decreases in strength with increasing population size. For a selectively neutral mutation, the expected change in gene frequency from one generation to the next is approximately the reciprocal of the population size. Motoo Kimura, the great modern population genetics theorist, formulated many results about the properties of neutral mutations. Consider an idealized diploid species of size N, where each individual possesses two copies of every gene. Therefore, there are 2N copies of each gene, and a newly arising mutation, which occurs as a unique mutation in a single offspring, will have initial frequency p = 1(/2N). Kimura proved, by diffusion approximation, that the probability of eventual fixation of a neutral mutation is its current frequency in the population. This means that for populations of large size, the probability that a neutral mutation will increase in frequency in a population and eventually completely replace the ancestral allele from which it arose, is very small. Most neutral mutations never become common, remaining rare in a population for a period of time before eventually ‘drifting’ to extinction. Kimura also showed that the expected time to fixation of a neutral mutation destined for fixation is 4N generations. For species such as insects, where population sizes must easily be in the millions, this means that neutral mutations destined for fixation spend a very long time segregating as genetic polymorphisms in populations. In contrast, a selectively favored mutation will be driven to fixation by positive natural selection much more quickly than a neutral mutation, and the time spent segregating in the population will be correspondingly shorter. This is one reason to suppose that at any given time in the history of a species, a large proportion of genetic variation will be selectively neutral. View chapterExplore book Read full chapter URL: Reference work2001, Brenner's Encyclopedia of Genetics (Second Edition)M. Kreitman Chapter Neutral Models of Genetic Drift and Mutation 2016, Encyclopedia of Evolutionary BiologyP.W. Messer Abstract Random genetic drift describes the stochastic fluctuations of allele frequencies due to random sampling in finite populations. Over time, genetic drift can lead to fixation or loss of genetic variants, thereby systematically eliminating diversity from a population. This trend is counterbalanced by mutations that continuously produce new variants. A number of powerful frameworks, such as coalescence theory, have been developed to study how these processes interact in shaping patterns of genetic diversity in populations. Random genetic drift and mutation also lie at the foundation of Kimura's neutral theory of evolution, which constitutes the standard null model of molecular population genetics. View chapterExplore book Read full chapter URL: Reference work2016, Encyclopedia of Evolutionary BiologyP.W. Messer Chapter Modern Morphometrics of Medically Important Insects 2011, Genetics and Evolution of Infectious DiseaseJean-Pierre Dujardin 16.3.2 Genetic Drift Since shape seems the output of a cascade of genes, it is expected that in natural conditions genetic drift will be a common factor of shape variation. Field observation has frequently observed significant shape differences between geographic areas (De la Riva et al., 2001; Dujardin et al., 2003; Gumiel et al., 2003; Dujardin and Le Pont, 2004a; Camara et al., 2006.; Aytekin et al., 2007; Henry et al., 2010). Laboratory experiments reproducing conditions favoring genetic drift between lines sharing the same environment were performed in Ae. aegypti. Using a set of three isofemale lines of Ae. aegypti monitored during 10 generations, a significant shift of shape appeared in one line, with nonsignificant changes in corresponding size (Jirakanjanakit et al., 2008). In this experiment, the change apparently produced by genetic drift did not affect the same landmarks as those affected by larval food or density variation (Jirakanjanakit et al., 2007). View chapterExplore book Read full chapter URL: Book2011, Genetics and Evolution of Infectious DiseaseJean-Pierre Dujardin Chapter Genetic Drift 2013, Brenner's Encyclopedia of Genetics (Second Edition)O. Honnay Genetic Drift and Gene Flow The loss of alleles through drift from a population, and the concomitantly increasing population genetic differentiation, can be counteracted if there is sufficient gene flow among populations. Gene flow can occur through the active migration of individuals or through the passive dispersal of seeds and pollen. It can be shown that for a large number of ideal populations: FST = 1/(4Nem + 1), where FST is a measure of the genetic differentiation between populations, m is the proportion of individuals that migrate between populations per generation, and Ne is the effective population size of the populations. This implies that one incoming migrant per generation in each population (Nem = 1) results in a population genetic differentiation of 0.20. This is considered as still acceptable and has been coined the ‘one-migrant-per-generation-rule’, which states that receiving one migrant per generation is sufficient to prevent genetic drift from reducing the population genetic variation and increasing the genetic differentiation. It should be noted that a genetic differentiation of 0.20 is a relatively arbitrary threshold, and that in the case of nonideal populations, much more migrants are required to counteract the effects of genetic drift. View chapterExplore book Read full chapter URL: Reference work2013, Brenner's Encyclopedia of Genetics (Second Edition)O. Honnay Related terms: Phylogenetics Population Genetics Linkage Disequilibrium Allele Frequency Haplotype Mitochondrial DNA Behavior (Neuroscience) Genetic Variation Neutral Gene Theory Intron View all Topics We use cookies that are necessary to make our site work. 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7432
https://fiveable.me/key-terms/criminology/social-structure-and-anomie
Social structure and anomie - (Criminology) - Vocab, Definition, Explanations | Fiveable | Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms Criminology Social structure and anomie 😈criminology review key term - Social structure and anomie Citation: MLA Definition Social structure and anomie refers to the concept that societal norms and values can become weakened or disrupted in times of social upheaval, leading to a breakdown in social order and increased deviance. This idea connects to how societal pressures can influence individual behavior, particularly when individuals feel disconnected from the goals and means of society, resulting in feelings of normlessness or anomie. 5 Must Know Facts For Your Next Test The concept of social structure and anomie was popularized by sociologist Robert K. Merton in the early 20th century as part of his Strain Theory. Merton argued that when society sets high aspirations without providing adequate means to achieve them, individuals may turn to deviant behaviors, reflecting a form of anomie. Anomie can occur during periods of rapid social change, economic instability, or crisis, as traditional norms and values become unclear or ineffective. Merton identified five modes of individual adaptation to anomie: conformity, innovation, ritualism, retreatism, and rebellion. Understanding social structure and anomie helps criminologists analyze how societal factors contribute to criminal behavior, emphasizing the importance of social context. Review Questions How does the concept of anomie relate to individual behavior within a changing social structure? Anomie relates to individual behavior by highlighting how societal changes can lead to feelings of normlessness, where individuals feel disconnected from cultural goals. When societal expectations are unclear or unreachable due to structural changes, people may adopt deviant behaviors as a way to cope or adapt. This breakdown in norms illustrates the impact of social structure on personal choices, making it essential to understand these dynamics in criminology. Discuss Merton's five modes of individual adaptation to anomie and their implications for understanding crime. Merton's five modes of adaptation include conformity, innovation, ritualism, retreatism, and rebellion. Each mode represents different responses individuals may have when facing anomie. For instance, those who innovate might engage in criminal behavior to achieve societal goals, while conformists adhere to societal rules despite challenges. Analyzing these adaptations allows criminologists to understand the various paths individuals may take in response to societal pressures and how these pathways can lead to criminal activities. Evaluate the role of social structure and anomie in the broader context of crime prevention strategies. Social structure and anomie play a critical role in shaping crime prevention strategies by emphasizing the need for stable norms and accessible means for achieving societal goals. By addressing structural inequalities and providing support for community cohesion, crime prevention efforts can reduce feelings of disconnection among individuals. Evaluating how social policies can strengthen community ties and ensure equal opportunities is vital for reducing crime rates and fostering a sense of belonging within society. Related terms Anomie:A state of normlessness or breakdown of social norms that occurs when individuals feel disconnected from societal goals and values. Social Disorganization Theory: A theory suggesting that a breakdown in social structures and relationships within communities leads to increased crime and deviance. Strain Theory:A theory that posits individuals may resort to deviant behavior when they experience a disconnect between culturally approved goals and the legitimate means available to achieve them. 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7433
https://asmedigitalcollection.asme.org/ebooks/book/135/ASME-Steam-Tables-Compact-Edition
ASME Steam Tables, Compact Edition | eBooks Gateway | ASME Digital Collection Skip to Main Content Open Menu Close Journals Open Menu ASME Journals Open Menu All Journals Virtual Issues Mechanical Engineering Magazine Select Articles Applied Mechanics Reviews ASCE-ASME Journal of Risk and Uncertainty in Engineering Systems, Part B: Mechanical Engineering ASME Letters in Dynamic Systems and Control ASME Letters in Translational Robotics ASME Open Journal of Engineering Journal of Applied Mechanics Journal of Autonomous Vehicles and Systems Journal of Biomechanical Engineering Journal of Computational and Nonlinear Dynamics Journal of Computing and Information Science in Engineering Journal of Dynamic Systems, Measurement, and Control Journal of Electrochemical Energy Conversion and Storage Journal of Electronic Packaging Journal of Energy Resources Technology Journal of Energy Resources Technology, Part A: Sustainable and Renewable Energy Journal of Energy Resources Technology, Part B: Subsurface Energy and Carbon Capture Journal of Engineering and Science in Medical Diagnostics and Therapy Journal of Engineering for Gas Turbines and Power Journal of Engineering for Sustainable Buildings and Cities Journal of Engineering Materials and Technology Journal of Fluids Engineering Journal of Heat and Mass Transfer Journal of Manufacturing Science and Engineering Journal of Mechanical Design Journal of Mechanisms and Robotics Journal of Medical Devices Journal of Micro and Nano Science and Engineering Journal of Nanotechnology in Engineering and Medicine Journal of Nondestructive Evaluation, Diagnostics and Prognostics of Engineering Systems Journal of Nuclear Engineering and Radiation Science Journal of Offshore Mechanics and Arctic Engineering Journal of Pressure Vessel Technology Journal of Solar Energy Engineering Journal of Thermal Science and Engineering Applications Journal of Tribology Journal of Turbomachinery Journal of Verification, Validation and Uncertainty Quantification Journal of Vibration and Acoustics ASTM Journals Open Menu All Journals Advances in Civil Engineering Materials (ACEM) Geotechnical Testing Journal (GTJ) Journal of Testing and Evaluation (JTE) Materials Performance and Characterization (MPC) Smart and Sustainable Manufacturing Systems (SSMS) Cement, Concrete and Aggregates (CCA) 1979-2004 Backfile Journal of ASTM International (JAI) 2004-2012 Backfile Journal of Composites, Technology & Research (JCTR) 1978-2003 Backfile Journal of Forensic Sciences (JOFS) 1972-2005 Backfile Conference Proceedings Open Menu ASME Conference Proceedings Open Menu All Conference Proceedings Browse by Series Browse by Subject Category Browse by Year eBooks Open Menu ASME eBooks ASTM eBooks Standards Open External Link Publishing Partners Open Menu ASME ASTM International Resources Open Menu About Authors Librarians FAQs Contact Us ASME.ORG Open External Link Purchase Open External Link Cart 0 User Tools Dropdown Cart 0 Sign In Open Menu Toggle Menu Menu Browse by Year Browse by Title Browse by Subject For Authors Open External Link About Open Menu About ASME eBooks Search Dropdown Menu header search search input Search input auto suggest filter your search Search Advanced Search Skip Nav Destination ASME Steam Tables, Compact Edition Available to Purchase By ASME Research and Technology, Committee on Water and Steam in Thermal Systems, Subcommittee on Properties of Steam ASME Research and Technology, Committee on Water and Steam in Thermal Systems, Subcommittee on Properties of Steam Search for other works by this author on: This Site PubMed Google Scholar ISBN-10: 079180254X No. of Pages: 32 DOI: Publisher: ASME Press Publication date: 2007 Description Properties of Saturated and Superheated Steam in U.S. Customary and SI Units from the IAPWS-IF97 International Standard for Industrial Use. This updated and concise booklet includes the following: U.S. Customary and SI Units Table 1. Saturated Water and Steam: Temperature Table Table 2. Saturated Water and Steam: Pressure Table Table 3. Superheated Steam (1 to 15,000 psia) SI Units Table 4 Saturated Water and Steam: Temperature Table Table 5. Saturated Water and Steam: Pressure Table Table 6. Superheated Steam (0.005 to 100 MPa) Unit Conversion Factors Mollier Diagrams (U.S. and SI units) Share Icon Share Facebook X LinkedIn Email Cite Icon Cite Permissions Search Site ASME Research and Technology, Committee on Water and Steam in Thermal Systems, Subcommittee on Properties of Steam ASME Steam Tables, Compact Edition. ASME Press, 2007. ISBN-10: 079180254X No. of Pages: 32 DOI: Publisher: ASME Press Published: 2007 Download citation file: Ris (Zotero) Reference Manager EasyBib Bookends Mendeley Papers EndNote RefWorks BibTex ProCite Medlars Purchase this Title $15.00 Purchase Learn about subscription and purchase options Product added to cart. Check OutContinue BrowsingClose Modal Table of Contents Front Matter Doi: Abstract View Chapter PDF Link PDF U.S. Customary and SI Units Doi: Abstract View Chapter PDF Link PDF Email alerts New eBook Alert Close Modal Copyright All rights reserved. Printed in the United States of America. 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7434
https://www.youtube.com/watch?v=5dtdupxyQK0
Write the equation of a circle given the center and a point it passes through Brian McLogan 1600000 subscribers 7669 likes Description 728011 views Posted: 12 May 2014 Learn how to write the equation of a circle. A circle is a closed shape such that all points are equidistance (equal distance) from a fixed point. The fixed point is called the center of the circle while the distance between any point of the circle and the center of the circle is called the radius of the circle. To write the equation of a circle, we need to know the length of the radius of the circle and the coordinate point of the center of the circle. Given a circle whose center is at (h, k) and the length of the radius is r, the equation of the circle is given by (x - h)^2 + (y - k)^2 = r^2. geometry #circles 352 comments Transcript: so what they're asking us to do in this equation is to write the equation of the circle so the main important thing to write the equation of the circle we have to know what is the general equation of circle which was in your notes last class period guo and that equation is x - h^ 2 + y - k^ 2 = R 2 so WR to write the equation of the circle we need to know the center and the radius right because remember the center is in form of HK and the radius is equal to R all right so we go and look at our general equation and we say all right are we provided what the center is or do we need to find it yeah we provided the center that's just H and K right perfect do we know the radius yeah no no we're not provided the radius right so in this problem we're not provided the radius however if I plugged in numbers for X and h and y and K would it make sense then I could solve for a variable R yes so I know H and K that's provided to us but it what could I use for X and Y four Zer a point on the circle which they give us right remember if you guys look at a circle graphed that's the center H comma K any point that I say is on the circle is a coordinate point which can represent XY the equation of the circle represents infinite many points amanii on the circle so therefore I can say that's XY so now all I need to do ladies and gentlemen is plug those points in and solve for R so X is 4 minus 5 squared + Y which is zero now here's where you got to be careful it's y - K right y minus what is k so it's going to be a positive all right equals R 2 well 4 - 5 is 1 1 2 is 1 plus uh -2 - 2 is going to or -2 -2 is pos2 pos4 equal R 2 5 = R 2 4 - 5 that's all it's going to be for that so 5 squar is there if I asked you what the radius was you could say the radius is the square root of five all right but we're not really talking about what R is we don't really care actually what the r is we just want to know what the equation of the circle is so the radius is square of five you square that and it's just equal to five so now that I know the center can I plug those in for H and K and now that I know what R square is can I plug that in yeah and that's the equation of my circle so therefore I have x - 5 2 + y + 2^ 2 = 5 notice how it's the opposite it's opposite H opposite K since H is positive it's xus h um since K is negative it's y plus two right so it's the opposite of those and then we figured out R squ which was five any other questions on that okay good talk
7435
https://link.springer.com/article/10.1007/s11165-024-10201-5
Exploring the Effect of Mathematics Skills on Student Performance in Physics Problem-Solving: A Structural Equation Modeling Analysis | Research in Science Education Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Log in Menu Find a journalPublish with usTrack your research Search Cart Search Search by keyword or author Search Navigation Find a journal Publish with us Track your research Home Research in Science Education Article Exploring the Effect of Mathematics Skills on Student Performance in Physics Problem-Solving: A Structural Equation Modeling Analysis Open access Published: 07 October 2024 Volume 55,pages 489–509, (2025) Cite this article Download PDF You have full access to this open access article Research in Science EducationAims and scopeSubmit manuscript Exploring the Effect of Mathematics Skills on Student Performance in Physics Problem-Solving: A Structural Equation Modeling Analysis Download PDF Tong Tong1, Feipeng Pi1, Siyan Zheng1, Yi Zhong1, Xiaochun Lin1& … Yajun WeiORCID: orcid.org/0000-0003-4871-96821 Show authors 5564 Accesses 4 Citations Explore all metrics Abstract Students’ success in physics problem-solving extends beyond conceptual knowledge of physics, relying significantly on their mathematics skills. Understanding the specific contributions of different mathematics skills to physics problem-solving can offer valuable insights for enhancing physics education. Yet such studies are rare, particularly at the high school level. This study addresses the underexplored area of this topic in secondary education by investigating the associations between physics problem-solving performance using a robust methodological framework. We applied exploratory factor analysis (EFA) to identify latent sub-mathmetics skills relevant to physics problem-solving and employed structural equation modeling (SEM) to examine the causal impact of these skills on students’ performance in physics. The study analyzed data from a municipal-wide assessment involving 1,878 grade 12 students in Southern China. The results demonstrate that mathematics skills impacting high school students’ physics problem-solving performance can be categorized into two sub skills, algebraic skills and geometric skills. It also indicates that algebraic skills have a stronger direct effect on physics problem-solving performance compared to geometric skills in high school setting. These findings suggest that integrating focused algebraic training within physics education can significantly improve student outcomes in STEM fields. We recommend that educators design curricula and instructional strategies that emphasize the development of algebraic skills necessary for solving complex physics problems. Additionally, these findings have important implications for policymakers, who should consider integrating targeted mathematics training within physics curricula to foster interdisciplinary learning and better prepare students for challenges in STEM education. Similar content being viewed by others A Mathematics Educator Walks into a Physics Class: Identifying Math Skills in Students’ Physics Problem-Solving Practices Article Open access 31 August 2023 The Integration of Mathematics in Physics Problem Solving: A Case Study of Greek Upper Secondary School Students Article 01 January 2016 Structural Skills of Students in Solving Physical–Mathematical Tasks Chapter© 2022 Explore related subjects Discover the latest articles, books and news in related subjects, suggested using machine learning. Education in Physics Mathematics Education Mathematical Teaching Educational Skills Mathematical Methods in Physics Mathematical Skills for Teaching Mathematics in Popular Science Use our pre-submission checklist Avoid common mistakes on your manuscript. Introduction In the landscape of STEM education, mathematics is often recognized as the foundational pillar that supports scientific learning (Dierdorp et al., 20141 "); Nakakoji & Wilson, 2018. First-year mathematics and its application to science: Evidence of transfer of learning to physics and engineering. Education Sciences, 8(1), 1–16. "); Turşucu et al., 2020. The effectiveness of activation of prior mathematical knowledge during problem-solving in physics. Eurasia Journal of Mathematics Science and Technology Education, 16(4), 1–24. ")). For example, physics problem-solving is not merely a function of conceptual understanding; it critically depends on students’ ability to apply mathematical principles to analyze and interpret physical phenomena. As such, mathematics serves as a crucial tool in navigating the quantitative aspects of problem-solving in physics (Dierdorp et al., 2014. Meaningful statistics in professional practices as a bridge between mathematics and science: An evaluation of a design research project. International Journal of STEM Education, 1, 1–15. ")). Despite the well-established interdependence of these disciplines, there remains a gap in our understanding of how specific mathematics skills—particularly geometric and algebraic skills—contribute to physics problem-solving performance, particularly at the secondary school level. Many studies have explored the impact of mathematics learning in science education, particularly in physics education. These studies often highlight a strong correlation between students’ mathematics skills and their performance in physics (Arbabifar, 2021b "); Dehipawala et al., 2014. Using mathematics review to enhance problem solving skills in general physics classes. In Proceedings of the 2014 Zone 1 Conference of the American Society for Engineering Education (pp. 1–4). IEEE. "); Matthews et al., 2009. Putting it into perspective: Mathematics in the undergraduate science curriculum. International Journal of Mathematical Education in Science and Technology, 40(7), 891–902. ")). Yet, further study on this topic is necessary for two major reasons. Firstly, much of this previous research has focused on university education settings but research findings at the university level might not automatically apply to a high school setting. (Jackson & Johnson, 2013. A hybrid model of mathematics support for science students emphasizing basic skills and discipline relevance. International Journal of Mathematical Education in Science and Technology, 44(6), 846–864. "); Meltzer, 2002. The relationship between mathematics preparation and conceptual learning gains in physics: A possible hidden variable in diagnostic pretest scores. American Journal of Physics, 70(12), 1259–1268. "); Rylands & Coady, 2009. Performance of students with weak mathematics in first-year mathematics and science. International Journal of Mathematical Education in Science and Technology, 40(6), 741–753. ")). The first research question in this study is to check if such a strong correlation, well documented at the college level, is evident among high school students, who are known to differ from adults in terms of factors influencing academic outcomes (Breitwieser & Brod, 2020. Cognitive prerequisites for generative learning: Why some learning strategies are more effective than others. Child Development, 92(1), 258–272. "); Dunlosky et al., 2013. Improving students’ learning with effective learning techniques: Promising directions from cognitive and educational psychology. Psychological Science in the Public Interest, 14(1), 4–58. "); Schneider & Preckel, 2017. Variables associated with achievement in higher education: A systematic review of meta-analyses. Psychological Bulletin, 143(6), 565. ")). Secondly, while previous research has documented correlations between general mathematics ability and science problem-solving, the specific roles of distinct mathematics skills—such as geometric and algebraic skills—remain underexplored, particularly in the context of secondary education. In other words, the second research question seeks to determine which skill, algebraic or geometric, has a higher impact, or if their impact levels are the same. Using data from a municipal-wide assessment in Southern China with 1,878 grade 12 students, we employ exploratory factor analysis (EFA) and structural equation modeling (SEM) to evaluate the contributions of these skills. Our findings indicate that algebraic skills have a more significant impact on physics problem-solving than geometric skills, highlighting the importance of incorporating targeted mathematics instruction within physics curricula. This approach can better equip students with the tools necessary to tackle complex scientific problems, ultimately enhancing STEM education outcomes. Literature Review The Relationship Between Mathematics and Physics Problem-Solving in Secondary Education Physics, being an elemental science, inherently involves problem-solving, which plays a crucial role in evaluating students’ knowledge and skills in physics education (Reddy & Panacharoensawad, 2017; Ince, 20181 ")). Successful problem-solving in physics is not solely dependent on students’ understanding of physics concepts but also significantly relies on their ability to apply mathematics principles (Redish & Kuo, 2015. Language of physics, Language of math: Disciplinary culture and dynamic epistemology. Science & Education, 24(5–6), 561–590. ")). Mathematics provides the language and tools necessary for the precise expression and application of physics laws and relationships, effectively bridging the gap between physics cognition and practical application (Bing & Redish, 2009. Analyzing problem solving using math in physics: Epistemological framing via warrants. Physical Review Physics Education Research, 5(2), 1–15. "); Franestian, 2020. Analysis problem solving skills of student in Junior High School. Journal of Physic: Conference Series, 1440(1), 1–5. ")). In this way, mathematics plays a vital role in enabling students to navigate the quantitative aspects of physics problem-solving, making it indispensable for achieving success in this field. Prior studies have established that students’ mathematics skills are closely associated with their performance in physics problem-solving. For example, Ogunleye (20119 ")) identified poor mathematics proficiency as a major obstacle to students’ ability to solve physics problems, while Panorkou and Germia (2021. Integrating math and science content through covariational reasoning: The case of gravity. Mathematical Thinking and Learning, 23(4), 318–343. ")) demonstrated that strong mathematics skills support a deeper understanding of physical phenomena, such as gravity, and enhance problem-solving abilities in this domain. These studies underscore the importance of integrating mathematics instruction with physics education to improve students’ problem-solving skills and overall academic performance. However, most research exploring the relationship between mathematics and physics has been conducted at the post-secondary level, focusing primarily on college students (Jackson & Johnson, 20139 "); Meltzer, 2002. The relationship between mathematics preparation and conceptual learning gains in physics: A possible hidden variable in diagnostic pretest scores. American Journal of Physics, 70(12), 1259–1268. "); Rylands & Coady, 2009. Performance of students with weak mathematics in first-year mathematics and science. International Journal of Mathematical Education in Science and Technology, 40(6), 741–753. ")). These studies have often highlighted correlations between general mathematics ability and success in physics courses (Arbabifar, 2021. Transfer of learning in a mathematical methods in physics course for undergraduate students of physics. European Journal of Physics, 42(4), 1–15. "); Dehipawala et al., 2014. Using mathematics review to enhance problem solving skills in general physics classes. In Proceedings of the 2014 Zone 1 Conference of the American Society for Engineering Education (pp. 1–4). IEEE. "); Matthews et al., 2009. Putting it into perspective: Mathematics in the undergraduate science curriculum. International Journal of Mathematical Education in Science and Technology, 40(7), 891–902. ")). While valuable, the findings from these studies may not be directly applicable to high school students, who differ from college students in significant ways (Talsma et al., 2018a. I believe, therefore i achieve (and vice versa): A meta-analytic cross-lagged panel analysis of self-efficacy and academic performance. Learning and Individual Differences, 61, 136–150. "); Schneider & Preckel, 2017. Variables associated with achievement in higher education: A systematic review of meta-analyses. Psychological Bulletin, 143(6), 565. ")). Teenagers are distinct from adult learners in several respects, including their cognitive processes, learning habits, emotional development, and self-efficacy (Richardson et al., 20128 "); Talsma et al., 2018a. I believe, therefore i achieve (and vice versa): A meta-analytic cross-lagged panel analysis of self-efficacy and academic performance. Learning and Individual Differences, 61, 136–150. ")). Also, teenager’s approach to learning and problem-solving is often less mature and more variable compared to that of college students, who have generally developed more advanced reasoning skills and learning habits (Schneider & Preckel, 2017. Variables associated with achievement in higher education: A systematic review of meta-analyses. Psychological Bulletin, 143(6), 565. ")). Consequently, the strategies and educational interventions that are effective for college students may not necessarily be effective for teenagers. This underscores the importance of studying the relationship between mathematics skills and physics problem-solving specifically within the high school context. By focusing on high school students, this study aims to provide insights that are directly relevant to secondary education, helping to bridge the gap between theory and practice in STEM education at this crucial stage of development. Latent Mathematics Skills Needed for Physics Problem-Solving When studying problem-solving tasks in physics, researchers and practitioners have identified several latent mathematics skills necessary for successfully solving these tasks (Awodun et al., 2013; Panorkou & Germia, 20217 "); Redish, 2023. Using Math in Physics: 6. Reading the physics in a graph. The Physics Teacher, 61(8), 651–656. "); Rebello et al., 2007. Transfer of learning in problem solving in the context of mathematics and physics. In D. H. Jonassen (Eds.), Learning to Solve Complex Scientific Problems, (pp. 223–246). Routledge. ")). Awodun et al. (2013. Mathematics skills as predictors of physics students’ performance in senior secondary schools. International Journal of Science and Research, 2(7), 391–394.")) explored the mathematics skills required to tackle physics questions among upper secondary school students, identifying six key skills: computation skills, geometry skills, graph and table interpretation skills, probability and statistics skills, algebraic skills, and measurement skills. Similarly, Daniel et al. (2020. Investigation of the role of mathematics on students’ performance in physics. Research in Science and Technological Education, 5(2), 101–108.")) concluded that deficiencies in analytical skills, algebraic processing skills, geometric skills, computational skills, as well as table and graph interpretation skills, were the primary reasons behind students’ poor performance in physics. Despite the recognition of these various sub-mathematics skills, little effort has been made to determine the relative importance of these skills. This gap in the literature calls for a deeper investigation into which sub-mathematics skill plays a more critical role in improving students’ physics problem-solving abilities. Algebraic Skills and Physics Problem-Solving Algebra is a fundamental component of mathematics and serves as a powerful tool in physics (Monk, 19945 ")). Algebraic skills in mathematics education include the ability to manipulate variables, solve equations, create formulas, work with functions, and apply these concepts to solve problems (Drijvers, 2011. Secondary algebra education: Revisiting topics and themes and exploring the unknown. Sense.")). Physics problem-solving benefits from the correct understanding and proper use of algebraic skills (Erdoğan et al., 2014. The effect of mathematical misconception on students’ success in kinematics teaching. Education Journal, 3(2), 90–94. ")). Research has shown that students with strong algebraic skills are more likely to succeed in physics problem-solving compared to those who struggle with algebra (Kanderakis, 2016. The mathematics of high school physics: Models, symbols, algorithmic operations and meaning. Science & Education, 25(7–8), 837–868. "); Rebello et al., 2007. Transfer of learning in problem solving in the context of mathematics and physics. In D. H. Jonassen (Eds.), Learning to Solve Complex Scientific Problems, (pp. 223–246). Routledge. ")). Awodun et al. (2013. Mathematics skills as predictors of physics students’ performance in senior secondary schools. International Journal of Science and Research, 2(7), 391–394.")) also found a positive impact of algebraic skills on problem-solving performance in secondary school physics. Geometric Skills and Physics Problem-Solving Geometric skills involve the ability to recognize and work with geometric shapes, visualize spatial relationships, sketch images, and apply geometric concepts to solve problems (Astuti et al., 20186 ")). According to Hoffer (1981. Geometry is more than proof. NCTM Journal, 74(1), 11–18. ")), geometric skills in mathematics encompass five types: visual skills, language skills, drawing skills, logical skills, and applied skills. These skills are essential in various physics problem-solving scenarios, such as analyzing forces, understanding light reflection and refraction, and evaluating projectile motion. The application of geometric skills is common in many areas of physics, and mastering these skills is vital for solving problems in this subject. Studies have consistently shown that geometric skills positively impact students’ performance in physics problem-solving at the secondary school level (Basson, 2002. Physics and mathematics as interrelated fields of thought development using acceleration as an example. International Journal of Mathematical Education in Science & Technology, 33(5), 679–690. "); Daniel et al., 2020. Investigation of the role of mathematics on students’ performance in physics. Research in Science and Technological Education, 5(2), 101–108.")). Other Skills and Physics Problem-Solving In addition to algebraic and geometric skills, graph and table interpretation skills, as well as probability and statistics skills, have been found to be closely related to students’ performance in physics (Awodun et al., 2013; Daniel et al., 2020). For instance, when analyzing a velocity-time graph, a student who cannot interpret the graph will struggle to determine whether an object is accelerating, decelerating, or moving at a constant speed. Research has consistently shown that graph and table interpretation plays a significant role in physics performance (McKenzie & Padilla, 19862 "); Redish, 2023. Using Math in Physics: 6. Reading the physics in a graph. The Physics Teacher, 61(8), 651–656. ")). Moreover, probability and statistics skills are closely tied to logical thinking in mathematics and encompass five key competencies: understanding probability concepts, applying these concepts, calculating probability values, interpreting these values accurately, and visualizing and communicating probabilities and their implications (Puspitasari et al., 2019. What are the difficulties in statistics and probability? Journal of Physics: Conference Series, 1402(7), 1–5. "); Tiro et al., 2021. Literacy description of probability for the senior secondary school students in makassar city. Journal of Physics: Conference Series, 1863(1), 1–9. ")). These skills are particularly relevant in the thermodynamics of physics; for example, students use them to grasp concepts like entropy and the second law from a microscopic perspective (Malgieri et al., 2016. Improving the connection between the microscopic and macroscopic approaches to thermodynamics in high school. Physics Education, 51(6), 1–13. ")). Research Questions In summary, previous research has identified various sub-mathematics skills required for high performance in physics problem-solving, but little effort has been made to explore which specific sub-mathemtaics skill plays a more significant role in enhancing physics learning. Furthermore, most studies examining the relationship between mathematics and physics problem-solving have focused on college students, leaving a gap in understanding how these skills function among high school students, who differ significantly from adult learners. This study aims to investigate the causal relationship between mathematics and physics problem-solving in upper secondary school physics education. Based on the data of a large scale assessment, we use exploratory factor analysis (EFA) to identify the latent mathematics skills involved in solving physics problems. Additionally, we employ structural equation modeling (SEM) to evaluate the relative strength of specific sub-mathematics skills’ impact on physics problem-solving for high school students. The specific research questions are as follows: Q1: Does the association between students’ mathematics skills and physics problem-solving performance also hold in a high school setting? Q2: What sub-mathematics skill plays a more important role in enhancing high school students’ physics problem-solving performance? Methodology The methodology of this study involved analyzing data from a municipal-wide assessment of 1,878 grade 12 students in Southern China, who were in the final year of a non-calculus-based physics course. The assessment included physics and mathematics tests, with the physics test comprising multiple-choice, blank-filling, and open-ended items, and the mathematics test featuring similar formats. Data analysis proceeded through three stages: assessing internal consistency and data collinearity, conducting exploratory factor analysis (EFA) to identify latent mathematics skills, and applying structural equation modeling (SEM) to explore the causal relationships between these skills and students’ performance in physics problem-solving. Sample The data were obtained from a well-developed municipal-wide assessment for evaluating upper secondary school students’ knowledge and problem-solving skills in physics. At the time of the test, the students were in the last year (grade 12) of a three-year-long physics course at upper secondary schools in a city in Southern China. The course, which is non-calculus-based, covers classical mechanics, electromagnetism, optics, and thermal physics, thereby preparing students for further studies in STEM subjects at university. These topics of physics are all covered in the first two years of the course, and the third year is solely dedicated to revision and training in physics problem-solving. 1,900 students were in grade 12 of the school, but 22 (1.16%) didn’t participate in either the physics or mathematics test of the assessment. Excluding these non-participants resulted in a valid sample size of 1,878. The data obtained from the school administration was anonymized, containing only a temporary test taker’s identification number (not real ID number) and performance scores on each item. The data did not contain students’ detailed demographic information, but the participants were nearly all aged 17 or 18 years, in accordance with the country’s strict age requirement for enrollment in K-12 education. Instruments The municipal-wide high school students’ knowledge and problem-solving skills physics assessment was structured with 10 multiple-choice items (7 single-correct-choice items, 3 multiple-correct-choice items), 2 blank-filling items, and 3 open-ended items. The assessment was a mock exam of the college entrance examination, with its content constructed by a panel of physics education experts and reviewed and validated by another team of experts independently. Table1 lists the measurement indicators of students’ physics knowledge for each item in this test. We scored the items dichotomously or polytomously based on the item formats. For the single-correct-choice items, which are multiple-choice problems with only one correct solution out of four options, students received a score of 1 if they answered the item correctly and a score of 0 otherwise. For the multiple-correct-choice items, which are multiple-choice problems with one to four possible correct solutions out of the list of four, students received a score of 2 if they picked all the correct options, a score of 1 if they picked some of the correct options, and a score of 0 if they picked any incorrect options. For the blank-filling items and open-ended items, we assigned a score of 1 to students whose item scores were higher than the average item score and a score of 0 to the rest, due to the discrete distribution. The Cronbach’s alpha of the 15 items ranged from 0.80 to 0.85 (Table1). The overall Cronbach’s alpha was 0.83, indicating the high reliability of the physics assessment. Table 1 The evaluation of students’ knowledge in physics Full size table The municipal-wide upper secondary school students’ mathematics skills assessment was structured with 11 multiple-choice items (8 single-correct-choice items, 3 multiple-correct-choice items), 3 blank-filling items, and 5 open-ended items. They were scored in the same way as the physics test. We scaled the scores of the students’ mathematics skills assessment to make the average assessment scores equal to 0 due to the discrete distribution. The overall Cronbach’s alpha of the assessment was 0.74, indicating the acceptable reliability of the physics assessment. Both the physics and mathematics assessments were part of a municipal-wide mock exam for the national college entrance examination, which is taken very seriously. All problem items were composed by a panel of six science and mathematics education experts, including university professors and experienced secondary school teachers, and were independently reviewed by another panel of such experts. The students took the 120-minute mathematics assessment on April 22, 2024, and the 75-minute physics assessment one day later. Data Analysis Almost all grade 12 students in the megalopolis municipal, with a population of 15 million, attend this municipal-wide mock exam in April 2024. Our data is extracted from one district of the municipality. This district has 1.5 million residents, with 1,900 grade 12 students who take science in high school. Among those 1,900 students, 1,878 successfully took both the mathematics and physics assessments. All of the 1,878 data points are included in this study. To answer the research questions, we implemented three stages to analyze the data. Firstly, we computed the covariances across items to evaluate the items’ internal consistency and data collinearity issues. Then, exploratory factor analysis (EFA) with eigenvalues and model fit was conducted to explore the latent factors of the mathematics skills in solving physics assessment items. The third stage was to conduct structural equation modeling (SEM) to estimate the causal relationships between students’ mathematics skills and their performance in physics problem-solving, given the latent mathematics skills Figure1. Fig. 1 Procedure of data analysis Full size image Exploring and Extracting Latent Factors Before implementing the EFA, we initially assessed the assumption of sphericity through Bartlett’s test (Bartlett, 19547 ")) to ensure that the correlation matrix was not random and evaluated sampling adequacy by the Kaiser-Meyer-Olkin (KMO test; Kaiser, 1974. An index of factorial simplicity. Psychometrika, 39, 31–36. ")) measure. According to widely accepted evaluation criteria (Khine et al., 2018. Students’ perceptions of the learning environment in tertiary science classrooms in Myanmar. Learning Environ Res, 21, 135–152. "); Tabachnick & Fidell, 2007. Using multivariate statistics (5th ed.). Pearson Education.")), EFA can proceed without violating assumptions or inflating estimated bias when the p-value of Bartlett’s test is less than 0.05, and the KMO test is greater than 0.5. Following the assumption evaluation, we structured EFA using the weighted least squares estimation and the promax oblique rotation (Hendrickson & White, 1964x ")), as we assumed the latent factors were correlated, and implemented three criteria to determine the number of latent factors. One criterion was based on the mathematical rules, as recommended by Cliff (1988. The eigenvalues-greater-than-one rule and the reliability of components. Psychological Bulletin, 103(2), 276–79. ")), indicating that the eigenvalues should be larger than 1. This criterion was conducted using the ‘psych’ package (Revelle, 2023. psych: Procedures for Psychological, Psychometric, and Personality Research (Version 2.3.6.) [Computer Software]. ")) in R version 4.3.1. The second criterion was based on the EFA model fit information, which included the chi-square test, Tucker-Lewis Index (TLI; Tucker & Lewis, 19730 ")), Bentler Comparative Fit Index (CFI; Bentler, 1990. Comparative fit indexes in structural models. Psychological Bulletin, 107(2), 238–246.")), Root Mean Square Error of Approximation (RMSEA; Steiger, 1990. Structural model evaluation and modification: An interval estimation approach. Multivariate Behavioral Research, 25(2), 173–180. ")), and standardized root mean square residual (SRMR). We extracted the number of latent factors when the chi-square p-value was smaller than 0.05, the CFI and TLI exceeded 0.95 (Hu & Bentler, 1999. Cutoff criteria for fit indexes in covariance structure analysis: Conventional criteria versus new alternatives. Structural Equation Modelling: A Multidisciplinary Journal, 6(1), 1–55. ")), and the RMSEA and SRMR were lower than 0.05 (Browne & Cudeck, 1993. Alternative way of assessing model fit. In K. A. Bollen & J. S. Long (Eds.), Testing structural equation models (pp. 136–162).")). These model fit indexes indicate a close fit of a model to data (Kline, 2016. Principles and practice of structural equation modeling (4th ed.). Guilford Press.")). This analysis was conducted using the ‘lavaan’ package (Rosseel, 2012. Lavaan: An R package for structural equation modeling. Journal of Statistical Software, 48(2), 1–36. ")) in R version 4.3.1. After extracting the number of factors, we named the latent factors based on the EFA factor loading outcomes. Additionally, items with factor loadings no less than 0.40 were selected as having a moderate or even strong association between the item and the latent factor (Stevens, 1992). Estimating Causal Relationships via SEM The SEM was structured in two parts. The first part involved the confirmatory factor analysis (CFA), which included a minimum of three items per factor to ensure the identification of model performance (Kline, 2016). The second part involved the structural part, including the path analysis for the direct effects in light of our hypothesis in the current study. The CFA was conducted using the ‘lavaan’ package (Rosseel, 20122 ")) with the method of maximum likelihood estimation. The model fit was evaluated by the chi-square test, Comparative Fit Index (CFI), Tucker-Lewis Index (TLI), Standardized Root Mean Square Residual (SRMR), and Root Mean Square Error of Approximation (RMSEA). The CFI and TLI exceeded 0.95 (Hu & Bentler, 1999. Cutoff criteria for fit indexes in covariance structure analysis: Conventional criteria versus new alternatives. Structural Equation Modelling: A Multidisciplinary Journal, 6(1), 1–55. ")), while the SRMR was below 0.05 (Hu & Bentler, 1999. Cutoff criteria for fit indexes in covariance structure analysis: Conventional criteria versus new alternatives. Structural Equation Modelling: A Multidisciplinary Journal, 6(1), 1–55. ")). Additionally, RMSEA was lower than 0.05 (Browne & Cudeck, 1993. Alternative way of assessing model fit. In K. A. Bollen & J. S. Long (Eds.), Testing structural equation models (pp. 136–162).")), indicating a perfect fit (Kline, 2016. Principles and practice of structural equation modeling (4th ed.). Guilford Press.")). For the path analysis, we estimated the direct effect of students’ mathematics skills on each extracted latent factor. The number of paths in the structural part was the same as the number of extracted factors. Results Covariances across Items Table2 shows the correlation between physics items and the mathematics assessment, none of which is smaller than 0.01. Therefore, we did not rule out any physics items in the following exploratory factor analysis. Table 2 The covariances across items Full size table Exploratory Factor Analysis The result of Bartlett’s test of sphericity was significant, with a p-value smaller than 0.001 ((\:{\chi\:}^{2})(105) = 5419.5514, p< .001) and the KMO test was 0.93. Both Bartlett’s test and the Kaiser-Meyer-Olkin measure indicated that the data were appropriate to proceed with the EFA. Figure2 is the scree plot visualizing the eigenvalues outcomes. The eigenvalues-greater-than-1 criterion suggested extracting two latent factors. The EFA model fit information with two factors, as shown in Table3, indicated that its TLI and CFI were larger than 0.950, and the RMSEA and SRMR were smaller than 0.050, all of which were acceptable values, indicating the extraction of two latent factors Figure3. Fig. 2 Scree plot of eigenvalues Full size image Fig. 3 Structural model with standardized estimates Note MS: Mathematics Skill; Gmt: Geometric Skills; Alg: Algebraic Skills Full size image Table 3 The EFA model fit information Full size table Table 4 The factor loadings of EFA Full size table Given the most salient manifest variables that latent factors have in common (Watkins, 20187 ")), the two factors were named as follows: Factor 1: Geometric Skills with three items and Factor 2: Algebraic Skills with seven items. Structural Equation Modeling Results The final fitted structural model had a perfect fit ((\:{\chi\:}^{2})(52) = 71.969, p< .05, CFI = 0.992, TLI = 0.991 SRMR = 0.039, and RMSEA = 0.014), as shown in Fig.2. Table5 lists the standardized factor loadings in the CFA part. The factor loadings of the 11 items ranged between 0.62 and 0.88, all of which were above 0.50. Each item had a statistically significant p-value smaller than 0.001. Table 5 The factor loadings of CFA Full size table In the structural part, there were two paths: students’ mathematics skills to algebraic skills in physics problem-solving and students’ mathematics skills to geometric skills in physics problem-solving. The standardized coefficient of the path from mathematics skills to the item responses involving algebraic skills was 0.797, indicating that students’ achievement increases by 0.797 scores in this category of physics problem-solving items as their mathematics skills increase by one unit. Likewise, the standardized coefficient of the path from mathematics skills to the item responses involving geometric skills was 0.737, indicating that students’ achievement increases by 0.737 scores in this category of physics problem-solving items as their mathematics skills increase by one unit. Discussion Interpretation of the Results This study aimed to investigate the causal effects of students’ mathematics skills on their physics problem-solving performance at the upper secondary school level. To achieve this, we employed Exploratory Factor Analysis (EFA) to identify the latent mathematics skills underlying physics problem-solving tasks. Subsequently, Structural Equation Modeling (SEM) was utilized to assess how these identified skills impacted students’ performance across different types of problem-solving tasks. This study reaffirms the strong relationship between mathematics skills and physics problem-solving performance observed in prior research. Our observation of the strong association between mathematics scores and physics scores in 1,878 students is consistent with many previous studies. For example, Arbabifar (2021b ")), Dehipawala et al. (2014. Using mathematics review to enhance problem solving skills in general physics classes. In Proceedings of the 2014 Zone 1 Conference of the American Society for Engineering Education (pp. 1–4). IEEE. ")), and Matthews et al. (2009. Putting it into perspective: Mathematics in the undergraduate science curriculum. International Journal of Mathematical Education in Science and Technology, 40(7), 891–902. ")) highlighted correlations between general mathematics ability and success in physics courses among college students. Similarly, Ogunleye (2011. Teachers and students perceptions of students problem-solving difficulties in physics: Implications for remediation. Journal of College Teaching & Learning, 6(7), 85–90. ")) identified poor mathematics proficiency as a major obstacle in students’ ability to solve physics problems, while Panorkou and Germia (2021. Integrating math and science content through covariational reasoning: The case of gravity. Mathematical Thinking and Learning, 23(4), 318–343. ")) demonstrated that strong mathematics skills support a deeper understanding of physical phenomena, such as gravity, and enhance problem-solving abilities in this domain. Our research confirms that mathematics skills significantly influence students’ ability to solve physics problems. Compared to college and university, physics education is sparsely researched at high school level (Kanim & Cid, 20206 ")). Yet previous studies suggest that teaching methods and educational research findings that are effective for college students might not automatically apply to teenagers, highlighting the need for more empirical studies to understand their learning (Breitwieser & Brod, 2020. Cognitive prerequisites for generative learning: Why some learning strategies are more effective than others. Child Development, 92(1), 258–272. "); Dunlosky et al., 2013. Improving students’ learning with effective learning techniques: Promising directions from cognitive and educational psychology. Psychological Science in the Public Interest, 14(1), 4–58. "); Schneider & Preckel, 2017. Variables associated with achievement in higher education: A systematic review of meta-analyses. Psychological Bulletin, 143(6), 565. ")). Our results from this group effectively bridge this critical gap in this topic of exploring the association between mathematics skills and physics problem solving, delivering insights essential for developing age-appropriate teaching strategies. Along with few studies targeted at teenage learners (Roorda, 2015. An actor-oriented transfer perspective on high school students’ development of the use of procedures to solve problems on rate of change. International Journal of Science and Mathematics Education, 13(4), 863–889. "); Turşucu, 2020), we extend this association from college students to high school teenage learners. Awodun et al. (2013) explored the underlying mathematics skills required to tackle physics questions among upper secondary school students, identifying six key skills: computation skills, geometry skills, graph and table interpretation skills, probability and statistics skills, algebraic skills, and measurement skills. Similarly, Daniel et al. (2020) concluded that deficiencies in analytical skills, algebraic processing skills, geometric skills, computational skills, as well as table and graph interpretation skills, were the primary reasons behind students’ poor performance in physics. Unlike these studies, our result indicates that mathematics skills influencing high school physics problem solving fall into only two categories, algebraic and geometric skills. We suggest this is because some other skills considered in those previous studies might potentially be included in either algebraic skills or geometric skills. For instance, computational skills may mainly be included in algebraic skills and graph interpretation skills may mainly fall into geometric skills. Also, some skills, such as analytical skill, are not subject-specific skill, as it doesn’t need mathematics conceptual knowledge and it is also an essential skill for other subjects (Powers & Enright, 19873 "); Bravo et al., 2016. Teaching for higher levels of thinking: Developing quantitative and analytical skills in environmental science courses. Ecosphere, 7(4). "); Wei, 2024. Recorded video versus narrated animation in teaching physics problem-solving: The influence of problem difficulty level. Journal of Baltic Science Education, 23(3), 570–587. ")). Furthermore, despite previous research has identified various sub-mathematics skills required for high performance in physics (Panorkou & Germia, 20217 "); Redish, 2023. Using Math in Physics: 6. Reading the physics in a graph. The Physics Teacher, 61(8), 651–656. "); Rebello et al., 2007. Transfer of learning in problem solving in the context of mathematics and physics. In D. H. Jonassen (Eds.), Learning to Solve Complex Scientific Problems, (pp. 223–246). Routledge. ")), little effort has been made to explore which specific sub-mathematics skill plays a more significant role in enhancing physics learning. Our findings clearly suggest the more critical role of algebraic skills in enhancing physics problem-solving abilities among high school students. This aspect of the research provides actionable insights that can guide curriculum developers and educators in refining their teaching strategies and educational content. Contribution to the Literature The contributions of this research to the current body of literature in this field are twofold. Firstly, by extending the exploration to high school students, this study enriches the sparse body of high school-level research (Kanim & Cid, 20206 ")), providing robust, data-driven evidence. Secondly, the study introduces an empirical comparison of sub-mathematics skills, marking the first time such a data-driven analysis of algebraic versus geometric skills has been conducted, to the best of our knowledge. In response to our research questions, the evidence clearly shows that an association between students’ mathematics skills and physics problem-solving performance does indeed hold in a high school setting. Furthermore, the evidence decisively indicates that algebraic skills are more critical than geometric skills for solving physics problems, underscoring the need for a curricular focus on enhancing algebraic training. These insights should compel policymakers to advocate for educational reforms that prioritize these skills. Educators and practitioners can leverage our findings to develop targeted instructional strategies, thereby better equipping students to tackle complex physics problems effectively. Teaching Strategies These results are particularly relevant when considering the implications for educational practices. Given the pronounced impact of algebraic skills on physics problem-solving, it is imperative that physics curricula include targeted algebra training. By systematically integrating algebraic problem-solving techniques into physics instruction, educators can better prepare students to navigate complex analytical tasks, thereby enhancing their overall performance. We suggest that educators adopt targeted strategies to enhance students’ problem-solving proficiency. Initially, educators should identify the specific mathematics skills required for each physics problem-solving task. They can then focus on strengthening these skills, helping students translate physics problems into mathematical terms and apply these concepts to various physical scenarios. Such preparation allows students to better understand and tackle problem-solving tasks by effectively using their knowledge of both mathematics and science. Furthermore, our findings indicate that mathematics skills significantly influence students’ success in physics problem-solving, particularly when tasks demand strong algebraic skills. For instance, when addressing Newton’s second law of motion, students who possess a strong foundation in algebraic manipulation—such as solving equations involving multiple variables—are significantly more likely to succeed. This illustrates the practical necessity of embedding algebraic training within the physics curriculum. To foster a practical integration of physics and mathematics, we propose that educators and policymakers implement a coordinated teaching approach, similar to the “Mathematical Methods in Physics” format (Arbabifar, 2021b ")). Such an approach would involve collaboration between mathematics and physics teachers to cover problem-solving tasks. For instance, in a scenario involving vehicle pursuits, mathematics teachers would focus on algebra, geometry, and data interpretation, while physics teachers would concentrate on the relevant physical principles, such as Newton’s laws. This collaborative teaching ensures that students receive a unified and coherent learning experience, enhancing their understanding and ability to apply mathematics skills in physics contexts. Additionally, we recommend enhancing communication and collaboration between mathematics and physics teachers to develop integrated teaching strategies. By forming instructional groups, educators can share insights and strategies for translating physics problems into mathematical language and vice versa. This approach not only helps teachers address their instructional challenges but also enriches the student learning experience by linking theoretical concepts with practical applications. Policy Making A review of education policies across various countries reveals a universal emphasis on the role of mathematics in science education (Department for Education, 2015; NGSS Lead States, 2013; MOE of PRC, 2020). For example, the Principles and Standards for School Mathematics highlight the necessity of mathematics education in preparing students for careers as engineers, scientists, and other related professions (NCTM, 2000). Similarly, the Next Generation Science Standards advocate for integrating rigorous scientific content with the mathematics skills commonly used by scientists and engineers (NGSS Lead States, 2013). This interdisciplinary approach is crucial for students’ overall development and equips them for future academic and professional challenges. In light of these insights and the findings of this research, we present specific recommendations for policymakers. Firstly, to better equip students for interdisciplinary challenges, we recommend the establishment of integrated mathematics and physics curricula. This approach ensures that students can apply mathematics skills in physics contexts with greater efficacy, reflecting the interconnected nature of these disciplines in both academic and professional environments. Secondly, teacher training programs for physics educators should focus on enhancing mathematics skills, especially algebraic skills. This aligns with Turşucu et al. (20206 ")), who suggest that the ability to explain fundamental mathematics should be a required competency for physics teachers. Accordingly, our study recommends a concentrated effort on developing algebraic skills in pre-service physics teachers to better prepare them for teaching complex physics problems. Implementing these strategies promises significant improvements in secondary school students’ abilities to solve physics problems, ultimately leading to enhanced outcomes in physics education and laying a robust foundation for their future endeavors. Limitations and Future Research While this study provides valuable insight into secondary school students’ mathematics skills and physics problem-solving performance, with practical implications for both practitioners and policymakers, it is important to acknowledge several limitations. Firstly, the physics problem-solving tasks in our empirical data encompassed multiple fields of physics knowledge, such as mechanics, thermodynamics, kinematics, and electromagnetism, rather than focusing on a single field. Each field may have varying degrees of association with mathematics skills, given the different latent mathematics skills involved in problem-solving tasks. In this study, we treated all physics knowledge as a whole to investigate the relationship between students’ problem-solving achievement and their mathematics skills, due to the limited number of items in our empirical data. Therefore, we suggest further refining the categorization of physics knowledge to gain a more comprehensive understanding of the specific impact of mathematics skills in different fields of physics, providing more targeted guidance for teaching practice. Secondly, the sample of the present study was limited to one city in southern China with a similar ethnic background. This limits our ability to generalize the findings to a broader population with different ethnic backgrounds or in other settings, such as rural schools. However, the problem-solving items in the empirical research are commonly included in the national physics assessment for upper secondary school students. Given that our sample was large enough to meet the requirements of a normal distribution, the identified sub-mathematics skills for physics problem-solving tasks are still informative. The conclusions of the current study, to some extent, provide valuable insights into teaching problem-solving in physics education in upper secondary schools. In future research, we recommend that researchers collect samples across multiple regions to draw more comprehensive and generalizable conclusions. Collecting samples from different countries for comparison, and combining the education policies of different countries for discussion, may help to draw more in-depth conclusions and promote the development of science education. Thirdly, the reliance on quantitative data may overlook the nuanced interplay between students’ cognitive, motivational, and contextual factors that can affect their performance in physics problem-solving. Qualitative data could enrich our understanding of how students apply their mathematics skills in physics contexts and the challenges they face in real-time problem-solving scenarios. Lastly, the study’s focus on the causal relationships between mathematics skills and physics problem-solving may not fully capture the dynamic and iterative processes that students undergo when learning and applying these skills. Future studies could benefit from longitudinal designs that track changes in students’ problem-solving capabilities and mathematics skills development over time, offering deeper insights into the learning trajectories and educational interventions that are most effective. Although the current study delved into the relationship between upper secondary school students’ mathematics skills and their performance in solving physics problems, the conclusions have the potential to be applied more broadly to science education. Other subjects within science, such as chemistry and biology, are also closely linked with mathematics. For example, in chemistry, mathematics is used to describe the relationship between conductivity and concentration (Shang, 20219 ")). Secondary school chemistry courses covering stoichiometry problems, Avogadro’s number, balancing equations, and other areas require a robust understanding of mathematics (Weisman, 1981. A mathematics readiness test for prospective chemistry students. Journal of Chemical Education, 58(7), 564. ")). In biology, mathematical methods are employed to understand the diversity and complexity of living systems (Kauffman, 1993. The origins of order: Self-organization and selection in evolution. Oxford University Press.")). As biology and biotechnology continue to evolve, there is an increasing demand for quantitative skills (Gross, 2004. Scientific illiteracy and the partisan takeover of biology. PLoS Biology, 4(5), 680–683. "); Karsai & Kampis, 2010. The crossroads between biology and mathematics: The scientific method as the basics of scientific literacy. BioScience, 60(8), 632–638. ")). This study’s methodological lens, focusing on the causal relationships facilitated by quantitative analysis through EFA and SEM, could serve as a model for similar research in other disciplines. By applying this proven approach, researchers and practitioners in fields like chemistry and biology could further investigate the pivotal role that mathematics plays in understanding and solving complex scientific problems. Therefore, we recommend that researchers and practitioners in other fields of science who share our research goal explore in-depth the causal relationship between mathematics ability and student achievement in problem-solving, considering various mathematics skills. Conclusions and Implications While prior studies exploring the relationship between mathematics skills and physics problem-solving at the high school level exist, they are rarer compared to those conducted at the college level. This study enriches this modest corpus by providing robust, data-driven evidence that highlights the distinct roles of algebraic and geometric skills in secondary physics education. Our analysis of the mathematics and physics assessment of 1878 grade 12 students using EFA and SEM demonstrates that mathematics skills needed for physics problem-solving in high school settings can be categorized into two sub-skills, with algebraic skills having a much more pronounced impact on physics problem-solving capabilities than geometric skills. These findings not only corroborate but also deepen our understanding of science education at this crucial academic stage. Implications for Educational Practice The significant influence of algebraic skills on physics problem-solving uncovered in this study suggests a need for strategic educational approaches. Physics educators are encouraged to integrate focused algebra training within their curricula, which could include the development of specific modules or interactive workshops that emphasize the application of algebraic concepts to physical problems. Such targeted training can equip students with the necessary tools to tackle complex scientific problems, thereby enhancing their overall academic success in STEM fields. Policy Recommendations This research underscores the importance of supportive educational policies that advocate for the integration of mathematics and physics education at the high school level. Policymakers should consider initiatives that fund and develop resources facilitating such integrated curricula. These policies could significantly contribute to creating a cohesive learning environment that effectively prepares students for advanced studies and careers in STEM disciplines. Conclusions This study significantly contributes to the limited but growing body of literature on high school students’ mathematics skills and their effect on physics problem-solving. 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ArticleGoogle Scholar Puspitasari, N., Afriansyah, E. A., Nuraeni, R., Madio, S. S., & Margana, A. (2019). What are the difficulties in statistics and probability? Journal of Physics: Conference Series, 1402(7), 1–5. ArticleGoogle Scholar Rebello, N. S., Cui, L., Bennett, A. G., Zollman, D. A., & Ozimek, D. J. (2007). Transfer of learning in problem solving in the context of mathematics and physics. In D. H. Jonassen (Eds.), Learning to Solve Complex Scientific Problems, (pp. 223–246). Routledge. Reddy, M. V. B., & Panacharoensawad, B. (2017). Students problem-solving difficulties and implications in physics: An empirical study on influencing factors. Journal of Education and Practice, 8(14). Redish, E. F. (2023). Using Math in Physics: 6. Reading the physics in a graph. The Physics Teacher, 61(8), 651–656. ArticleGoogle Scholar Redish, E. F., & Kuo, E. (2015). Language of physics, Language of math: Disciplinary culture and dynamic epistemology. Science & Education, 24(5–6), 561–590. 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International Journal of Mathematical Education in Science and Technology, 40(6), 741–753. ArticleGoogle Scholar Schneider, M., & Preckel, F. (2017). Variables associated with achievement in higher education: A systematic review of meta-analyses. Psychological Bulletin, 143(6), 565. ArticleGoogle Scholar Shang, H. (2021). Connecting chemistry to mathematics by establishing the relationship between conductivity and concentration in an interdisciplinary, computer-based project for high school chemistry students. Journal of Chemical Education, 98(3), 796–804. ArticleGoogle Scholar Steiger, J. H. (1990). Structural model evaluation and modification: An interval estimation approach. Multivariate Behavioral Research, 25(2), 173–180. ArticleGoogle Scholar Stevens, J. (1992). Applied multivariate statistics for the social sciences (2nd ed.). Lawrence Erlbaum Associates. Tabachnick, B. G., & Fidell, L. S. (2007). Using multivariate statistics (5th ed.). Pearson Education. Talsma, K., Schüz, B., Schwarzer, R., & Norris, K. (2018). Miscalibration of self-efficacy and academic performance: Self-efficacy ≠ self-fulfilling prophecy. Learning and Individual Differences, 69, 182–195. ArticleGoogle Scholar Talsma, K., Schüz, B., Schwarzer, R., & Norris, K. (2018a). I believe, therefore i achieve (and vice versa): A meta-analytic cross-lagged panel analysis of self-efficacy and academic performance. Learning and Individual Differences, 61, 136–150. ArticleGoogle Scholar Tiro, M. A., Ruliana, & Aswi, A. (2021). Literacy description of probability for the senior secondary school students in makassar city. Journal of Physics: Conference Series, 1863(1), 1–9. Tucker, L. R., & Lewis, C. (1973). A reliability coefficient for maximum likelihood factor analysis. Psychometrika, 38(1), 1–10. ArticleGoogle Scholar Turşucu, S., Spandaw, J., & de Vries, M. J. (2020). The effectiveness of activation of prior mathematical knowledge during problem-solving in physics. Eurasia Journal of Mathematics Science and Technology Education, 16(4), 1–24. ArticleGoogle Scholar Watkins, M. W. (2018). Exploratory factor analysis: A guide to best practice. Journal of Black Psychology, 44(3), 219–246. ArticleGoogle Scholar Wei, Y., Chen, X., Zhong, Y., Liu, G., Wang, M., Pi, F., & Li, C. (2024). Recorded video versus narrated animation in teaching physics problem-solving: The influence of problem difficulty level. Journal of Baltic Science Education, 23(3), 570–587. ArticleGoogle Scholar Weisman, R. L. (1981). A mathematics readiness test for prospective chemistry students. Journal of Chemical Education, 58(7), 564. ArticleGoogle Scholar Download references Funding No funding was received to assist with the preparation of this manuscript. Author information Authors and Affiliations School of Physics and Materials Science, Guangzhou University, Building of Administration West 401, No. 230, West Waihuan Street, Higher Education Mega Center, Panyu District, Guangzhou City, Guangdong Province, China Tong Tong,Feipeng Pi,Siyan Zheng,Yi Zhong,Xiaochun Lin&Yajun Wei Authors 1. Tong TongView author publications Search author on:PubMedGoogle Scholar 2. Feipeng PiView author publications Search author on:PubMedGoogle Scholar 3. Siyan ZhengView author publications Search author on:PubMedGoogle Scholar 4. Yi ZhongView author publications Search author on:PubMedGoogle Scholar 5. Xiaochun LinView author publications Search author on:PubMedGoogle Scholar 6. Yajun WeiView author publications Search author on:PubMedGoogle Scholar Corresponding author Correspondence to Yajun Wei. Ethics declarations Competing Interests The authors declare that they have no competing interests. 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To view a copy of this licence, visit Reprints and permissions About this article Cite this article Tong, T., Pi, F., Zheng, S. et al. Exploring the Effect of Mathematics Skills on Student Performance in Physics Problem-Solving: A Structural Equation Modeling Analysis. Res Sci Educ55, 489–509 (2025). Download citation Accepted: 18 September 2024 Published: 07 October 2024 Issue Date: June 2025 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Keywords Physics problem-solving Mathematics skills Secondary education Structural equation modeling Use our pre-submission checklist Avoid common mistakes on your manuscript. Sections Figures References Abstract Introduction Literature Review Latent Mathematics Skills Needed for Physics Problem-Solving Algebraic Skills and Physics Problem-Solving Geometric Skills and Physics Problem-Solving Other Skills and Physics Problem-Solving Methodology Sample Results Discussion Teaching Strategies Policy Recommendations Conclusions Data availability References Funding Author information Ethics declarations Additional information Rights and permissions About this article Advertisement Fig. 1 View in articleFull size image Fig. 2 View in articleFull size image Fig. 3 View in articleFull size image Arbabifar, F. (2021). Transfer of learning in a mathematical methods in physics course for undergraduate students of physics. European Journal of Physics, 42(4), 1–15. ArticleGoogle Scholar Astuti, R., Suryadi, D., & Turmudi (2018). Analysis on geometry skills of junior high school students on the concept congruence based on Van Hiele’s geometric thinking level. 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ArticleGoogle Scholar Puspitasari, N., Afriansyah, E. A., Nuraeni, R., Madio, S. S., & Margana, A. (2019). What are the difficulties in statistics and probability? Journal of Physics: Conference Series, 1402(7), 1–5. ArticleGoogle Scholar Rebello, N. S., Cui, L., Bennett, A. G., Zollman, D. A., & Ozimek, D. J. (2007). Transfer of learning in problem solving in the context of mathematics and physics. In D. H. Jonassen (Eds.), Learning to Solve Complex Scientific Problems, (pp. 223–246). Routledge. Reddy, M. V. B., & Panacharoensawad, B. (2017). Students problem-solving difficulties and implications in physics: An empirical study on influencing factors. Journal of Education and Practice, 8(14). Redish, E. F. (2023). Using Math in Physics: 6. Reading the physics in a graph. The Physics Teacher, 61(8), 651–656. ArticleGoogle Scholar Redish, E. F., & Kuo, E. (2015). Language of physics, Language of math: Disciplinary culture and dynamic epistemology. Science & Education, 24(5–6), 561–590. ArticleGoogle Scholar Revelle, W. (2023). psych: Procedures for Psychological, Psychometric, and Personality Research (Version 2.3.6.) [Computer Software]. Richardson, M., Abraham, C., & Bond, R. (2012). Psychological correlates of university students’ academic performance: A systematic review and meta-analysis. Psychological Bulletin, 138(2), 353–387. ArticleGoogle Scholar Roorda, G., Vos, P., & Goedhart, M. J. (2015). An actor-oriented transfer perspective on high school students’ development of the use of procedures to solve problems on rate of change. International Journal of Science and Mathematics Education, 13(4), 863–889. ArticleGoogle Scholar Rosseel, Y. (2012). Lavaan: An R package for structural equation modeling. Journal of Statistical Software, 48(2), 1–36. ArticleGoogle Scholar Rylands, L. J., & Coady, C. (2009). Performance of students with weak mathematics in first-year mathematics and science. International Journal of Mathematical Education in Science and Technology, 40(6), 741–753. ArticleGoogle Scholar Schneider, M., & Preckel, F. (2017). Variables associated with achievement in higher education: A systematic review of meta-analyses. Psychological Bulletin, 143(6), 565. ArticleGoogle Scholar Shang, H. (2021). Connecting chemistry to mathematics by establishing the relationship between conductivity and concentration in an interdisciplinary, computer-based project for high school chemistry students. Journal of Chemical Education, 98(3), 796–804. ArticleGoogle Scholar Steiger, J. H. (1990). Structural model evaluation and modification: An interval estimation approach. Multivariate Behavioral Research, 25(2), 173–180. ArticleGoogle Scholar Stevens, J. (1992). Applied multivariate statistics for the social sciences (2nd ed.). Lawrence Erlbaum Associates. Tabachnick, B. G., & Fidell, L. S. (2007). Using multivariate statistics (5th ed.). Pearson Education. Talsma, K., Schüz, B., Schwarzer, R., & Norris, K. (2018). Miscalibration of self-efficacy and academic performance: Self-efficacy ≠ self-fulfilling prophecy. Learning and Individual Differences, 69, 182–195. ArticleGoogle Scholar Talsma, K., Schüz, B., Schwarzer, R., & Norris, K. (2018a). I believe, therefore i achieve (and vice versa): A meta-analytic cross-lagged panel analysis of self-efficacy and academic performance. Learning and Individual Differences, 61, 136–150. ArticleGoogle Scholar Tiro, M. A., Ruliana, & Aswi, A. (2021). Literacy description of probability for the senior secondary school students in makassar city. Journal of Physics: Conference Series, 1863(1), 1–9. Tucker, L. R., & Lewis, C. (1973). A reliability coefficient for maximum likelihood factor analysis. Psychometrika, 38(1), 1–10. ArticleGoogle Scholar Turşucu, S., Spandaw, J., & de Vries, M. J. (2020). The effectiveness of activation of prior mathematical knowledge during problem-solving in physics. Eurasia Journal of Mathematics Science and Technology Education, 16(4), 1–24. ArticleGoogle Scholar Watkins, M. W. (2018). Exploratory factor analysis: A guide to best practice. Journal of Black Psychology, 44(3), 219–246. ArticleGoogle Scholar Wei, Y., Chen, X., Zhong, Y., Liu, G., Wang, M., Pi, F., & Li, C. (2024). Recorded video versus narrated animation in teaching physics problem-solving: The influence of problem difficulty level. Journal of Baltic Science Education, 23(3), 570–587. ArticleGoogle Scholar Weisman, R. L. (1981). A mathematics readiness test for prospective chemistry students. Journal of Chemical Education, 58(7), 564. 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https://www.youtube.com/watch?v=wzQstigxbuo
The Distance Formula: Finding the Distance Between Two Points Professor Dave Explains 3890000 subscribers 1277 likes Description 104561 views Posted: 11 Nov 2017 One thing we may want to do in algebra is compute the distance between two points on the coordinate plane. Luckily, there is a handy dandy equation for doing just that! Let's derive it first, and then practice using it. Watch the whole Mathematics playlist: Classical Physics Tutorials: Modern Physics Tutorials: General Chemistry Tutorials: Organic Chemistry Tutorials: Biochemistry Tutorials: Biology Tutorials: EMAIL► ProfessorDaveExplains@gmail.com PATREON► Check out "Is This Wi-Fi Organic?", my book on disarming pseudoscience! Amazon: Bookshop: Barnes and Noble: Book Depository: 61 comments Transcript: It’s Professor Dave, let’s learn the distance formula. As we navigate the coordinate plane in our algebraic adventures, one thing we might want to do is find the distance between two points. Luckily, we are well equipped to do this, since we know the Pythagorean Theorem. If you need to review it, check out the tutorials on how to use the theorem, as well as how the theorem was derived, in the geometry portion of this playlist. Otherwise, let’s just select two points in the coordinate plane, and draw a line between them. The length of this line is what we are trying to figure out, and in order to get it, we can make a right triangle. This line will be the hypotenuse, and the legs will be the horizontal and vertical line segments we need to construct the triangle. Now let’s label the hypotenuse as D, the distance we want to know. The legs can be expressed as the difference between the respective points that mark their ends. This horizontal line segment has a length of X two minus X one, since it only moves in the X direction. The vertical line segment has a length of Y two minus Y one, since it only moves in the Y direction. Now, we just use the Pythagorean Theorem. The square of this leg plus the square of the other leg equals the hypotenuse squared, or D squared. Then, all we do is take the square root of both sides to solve for D. We will only take the positive value here, as negative length doesn’t make any sense. And there you have it, we have derived the distance formula. Let’s give it a shot. Take the points negative two, two, and two, five. What is the distance covered by the line segment that connects these two points? Well, let’s take our X values and our Y values, and plug them into the formula. That gives us four, which squared is sixteen, and three, which squared is nine. Add them up and we get twenty-five, take the square root, and that gives us five for the distance. This is the same value we would get if we contructed a triangle and performed the Pythagorean Theorem. So the distance formula is a generalized way to simply plug in the coordinates of two points, and find the distance between them. Let's check comprehension.
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https://www.cda.gov.sg/professionals/diseases/marburg-virus-disease/
Skip to main content A Singapore Government Agency Website Official website links end with .gov.sg Government agencies communicate via .gov.sg websites (e.g. go.gov.sg/open). Trusted websites Secure websites use HTTPS Look for a lock () or https:// as an added precaution. Share sensitive information only on official, secure websites. News and events Resources Home Professionals Diseases Marburg virus disease Marburg virus disease Marburg virus (MARV) and Ravn virus (RAVV) Last updated 26 March 2025 On this page Overview Disease epidemiology Pathogen(s) Transmission Clinical features Risk factors Diagnosis Treatment and management Precaution, prevention, and control Notification Resources Overview Marburg Virus Disease (MVD), formerly known as Marburg haemorrhagic fever, is a rare but severe zoonotic disease. It is a rapidly progressive febrile illness that leads to haemorrhagic shock and death in a large proportion of cases. Disease epidemiology MVD was initially detected in 1967 when two simultaneous outbreaks occurred in Germany and Serbia, arising from laboratory researchers who were inadvertently exposed to the virus after performing necropsies on infected non-human primates that were imported from Africa for use in vaccine production. Since then, there have been several MVD outbreaks, all of which occurred in Central and East Africa, primarily in Uganda. The largest MVD outbreak to date involved more than 370 confirmed cases in Angola in 2004 and was centred on a paediatric ward where the infection apparently spread through contaminated transfusion equipment. There were also reports on sporadic cases of international travellers developing MVD after visiting caves colonised by fruit bats in Uganda. In the first quarter of 2023, MVD outbreaks had been declared in Equatorial Guinea on 13 February, and in Tanzania on 21 March. Both outbreaks are geographically distinct, occurring on opposite sides of the African continent. The source of infection in both outbreaks is unknown – there were no known epidemiological links between the two outbreaks. As of 2 and 8 June 2023, the outbreaks were declared over in Tanzania and Equatorial Guinea respectively. The case fatality ratios (CFR) have varied from 23% to 90% in past MVD outbreaks. Another outbreak was confirmed in Rwanda on 27 September 2024 and was declared over by WHO on 20 December 2024. A total of 66 confirmed cases and 15 deaths were recorded. On 14 January 2025, the WHO reported a suspected outbreak of MVD in the Kagera region of Tanzania. Following investigations, on 20 January 2025, the outbreak was confirmed, making it the latest MVD outbreak. As of 31 January 2025, two confirmed cases and eight probable cases had been reported. Pathogen(s) Marburg virus (MARV) and Ravn virus (RAVV) of the species Orthomarburgvirus marburgense are the causative agents of Marburg virus disease (MVD). Both viruses are part of the Filoviridae family (filovirus) to which Orthoebolavirus genus belongs. Though caused by different viruses, Ebola and Marburg diseases are clinically similar. Egyptian fruit bats (Rousettus aegyptiacus) native to Africa are considered the natural hosts. Transmission Transmission can occur through direct contact with body fluids of infected persons or animals, or indirectly via contaminated surfaces and materials. Animal-to-human transmission of the virus occurs through close contact with contaminated tissues and bodily fluids of infected animals. This may occur during prolonged exposure to mines or caves inhabited by Rousettus bat colonies. Other animal-to-human transmission may occur from contact or consumption of tissue and bodily fluids from infected Non-Human-Primates (NHPs) without appropriate protective equipment or thorough cooking. Human-to-human transmission can occur through direct contact with blood, bodily fluids, or secretions from an infected person through non-intact skin, mucosa, percutaneous injury, or sexual intercourse. Contact with infected corpses or environments that have been contaminated with an MVD patient’s bodily fluids may also spread the infection. Incubation period: Typically 5 to 10 days; range is 2 to 21 days. Infectious period: Infected individuals are not contagious during the incubation period and become infectious once they begin to develop symptoms, and remain infectious as long as their blood and body fluids contain the virus. Marburg Virus may also be transmitted through semen after recovery from MVD. There is no evidence that Marburg virus can spread through sex or other contact with vaginal fluids from a woman who has recovered from MVD. Clinical features Presentation of MVD is similar to the Ebola Virus Disease. Early symptoms such as fever, headache, sore throat, myalgia, joint pain, and malaise can be varied and non-specific. Progression to maculopapular rash, abdominal pain, nausea, vomiting, diarrhoea, gum bleeding and gastrointestinal bleeding should alert the clinician to MVD. In the late phase of the disease, patients may experience orchitis and central nervous system (CNS) manifestations of confusion, agitation, seizures or coma. Risk factors Risk factors include exposure to: African fruit bats, or their saliva, urine or excretions Infected non-human primates, including handling of bush meat Blood and body fluids from infected persons, or contaminated environment and materials Diagnosis Diagnosis in acute infection is by detection of MARV by polymerase chain reaction (PCR) in blood specimens. Diagnostic tests for MVD should be performed only after discussion with MOH. All blood samples taken from inpatients are to be sent to the National Public Health Laboratory (NPHL) for MVD testing through the laboratory. NPHL will provide laboratory support and guidelines for testing, including protocols for specimen transport, testing and reporting. Treatment and management To date, there are no approved vaccines or antiviral treatments for MVD. Vaccines and treatments developed against Ebola virus disease are ineffective against MVD. The World Health Organization (WHO) is prepared to support trials of four Marburg vaccines in Equatorial Guinea and Tanzania. In the absence of proven antivirals, supportive care remains the mainstay of treatment for MVD. This includes: Diagnosis and treatment of concomitant infections such as malaria or bacteremia Active management of fluids and electrolytes Managing nausea and vomiting Supplemental oxygen (if needed) Inotropic support (if needed) Analgesics (if needed) Blood product support if haemorrhage occurs Support for multi-organ dysfunction, e.g., renal replacement therapy (if needed) Precaution, prevention, and control General advice All suspect cases of MVD should be isolated immediately to limit contact with other persons while awaiting risk assessment, conveyance and admission. Strict contact and airborne precautions should be exercised. Staff are reminded to practice good hand hygiene at all times. For a suspect case with: Low index of suspicion for MVD, the patient will be admitted and isolated in an airborne infection isolation room2 (AIIR) in isolation wards in the respective acute care hospital. This includes suspect cases presenting at emergency department or specialised outpatient clinics. High index of suspicion for MVD or confirmed to have MVD, arrangements will be made to transfer the patient using a trained ambulance operator to NCID to be managed at the high-level isolation unit. All cases with a high index of suspicion of MVD should be isolated in Airborne Infection Isolation Room (AIIR) at a minimum and placed on strict contact and airborne isolation precautions. Only essential personnel trained in donning/doffing of appropriate PPE should attend to a suspect or confirmed MVD case or their body fluids. All personnel attending to individuals with: High index of suspicion for MVD or confirmed to have MVD should don PPE comprising: Disposable fluid-resistant hood to cover the head and neck areas (staff with long hair may wish to use head cover prior to wearing hood) Eye protection gear (e.g., disposable downward face shields secured at forehead or goggles) Fluid-repellent N95 mask Inner and outer disposable fluid-resistant (AAMI 4) gown (should extend at least until the ankle portion of the boot cover) 12-inch double nitrile or latex gloves certified for healthcare use (extended cuffs should reach up to mid forearm) Disposable fluid resistant boot covers (should extend up to knee-height) Low index of suspicion for MVD should minimally don PPE comprising: Fluid-repellent N95 mask Eye protection gear (e.g., disposable downward face shields secured at forehead or goggles) Disposable fluid-resistant (AAMI 4) gown 12-inch nitril or latex gloves certified for healthcare use Staff attending to suspect MVD case should be closely monitored (e.g. twice daily temperature monitoring), even if they wear the appropriate PPE. Unprotected exposure should be reported to MOH immediately. A log of all staff attending to the patient should be maintained. Persons with skin or mucousal exposure to body fluids from a suspect MVD case should immediately wash the affected skin surfaces with soap and water. Mucous membranes (e.g. conjunctiva) should be irrigated with copious amounts of water or eyewash solution. Notification MVD is an emerging infectious disease and is legally notifiable in Singapore. All suspected cases of MVD should be reported to MOH immediately and instructions will be provided on further management of these cases. MOH will arrange for transfer of suspected cases to the NCID on a case-by-case basis. All confirmed MVD cases will be managed at the HLIU in NCID. Who should notify: Medical Practitioners Laboratories When to notify: On clinical suspicion Suspect case definition: Fever (>38°C) or current history of fever Onset of symptoms within 21 days of: Travel history to any area or region with confirmed case(s) of MVD [refer to World Health Organization (WHO)] Close contact to a confirmed or suspect case of MVD On laboratory confirmation of positive PCR How to notify: Please refer to the Infectious Disease Notification for more information. Notification timeline: Immediately Resources Get MVD situation updates from WHO. Back to top
7438
https://www.bbc.co.uk/bitesize/topics/zr269ty/watch/zrgctcw
Dividing by decimals - Number: Video playlist - BBC Bitesize BBC Homepage Skip to content Accessibility Help Sign in Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds More menu More menu Search Bitesize Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds Close menu Bitesize Menu Home Learn Study support Careers Teachers Parents Trending My Bitesize More England Early years KS1 KS2 KS3 GCSE Functional Skills Northern Ireland Foundation Stage KS1 KS2 KS3 GCSE Scotland Early Level 1st Level 2nd Level 3rd Level 4th Level National 4 National 5 Higher Core Skills An Tràth Ìre A' Chiad Ìre An Dàrna Ìre 3mh ìre 4mh ìre Nàiseanta 4 Nàiseanta 5 Àrd Ìre Wales Foundation Phase KS2 KS3 GCSE WBQ Essential Skills Cyfnod Sylfaen CA2 CA3 CBC TGAU International KS3 IGCSE More from Bitesize About us All subjects All levels Primary games Secondary games GCSE Edexcel Number: Video playlist Part ofMathsNumber Now playing video 2 of 10 Dividing by decimals Description A GCSE Maths video about dividing numbers by decimals. The video shows how to use equivalent fractions to divide numbers by decimals. Find out more about GCSE Maths Number Now playing video 2 of 10 Adding and subtracting with negative numbers. Video Adding and subtracting with negative numbers 1 of 10 3:05 Now playing. Dividing by decimals. Video Dividing by decimals 2 of 10 Now playing 2:53 Up next. Converting recurring decimals. Video Converting recurring decimals 3 of 10 Up next 3:26 Rounding to significant figures. Video Rounding to significant figures 4 of 10 2:34 Upper and lower bounds. Video Upper and lower bounds 5 of 10 2:54 Dividing by fractions. Video Dividing by fractions 6 of 10 1:56 Laws of indices. Video Laws of indices 7 of 10 2:26 Negative indices. Video Negative indices 8 of 10 3:14 Converting between an ordinary number and standard form. Video Converting between an ordinary number and standard form 9 of 10 3:16 Calculating in standard form. Video Calculating in standard form 10 of 10 2:54 Language: Home News Sport Earth Reel Worklife Travel Culture Future Music TV Weather Sounds Terms of Use About the BBC Privacy Policy Cookies Accessibility Help Parental Guidance Contact the BBC BBC emails for you Advertise with us Do not share or sell my info Copyright © 2025 BBC. The BBC is not responsible for the content of external sites. Read about our approach to external linking.
7439
https://www.nv5geospatialsoftware.com/docs/BackgroundFrameLineCentralProjections.html
Frame and Line Central Projections NV5 GEOSPATIAL SOFTWARE Docs Center IDL Programming IDL Reference Using IDL Modules Advanced Math and Stats Dataminer DICOM Toolkit Libraries Astronomy Library Beaumont Library Buie Library Catalyst Library Coyote Library Dáithí Library Galloy Library JBIU Library JHUAPLS1R Library Mankoff Library Markwardt Library Motley Library Salvaggio Library ENVI Using ENVI Modules Atmospheric Correction Crop Science Deep Learning DEM Extraction Feature Extraction Machine Learning NITF Photogrammetry ENVI API Resources Licensing Platform Support [x] Docs Center IDL Programming IDL Programming- [x] IDL Reference Using IDL Modules Modules- [x] Advanced Math and Stats Dataminer DICOM Toolkit Libraries Libraries- [x] Astronomy Library Beaumont Library Catalyst Library Coyote Library Dáithí Library Galloy Library JBIU Library JHUAPLS1R Library Mankoff Library Markwardt Library Motley Library Salvaggio Library ENVI ENVI- [x] Using ENVI Modulues Modules- [x] Atmospheric Correction Crop Science Deep Learning DEM Extraction Feature Extraction NITF Photogrammetry ENVI API Resources Resources- [x] Using ENVI Modulues Modules- [x] Licensing PDF Guides Platform Support ENVI API >Docs Center>Using ENVI>Frame and Line Central Projections ### Frame and Line Central Projections Loading… Search Results Frame and Line Central Projections Background ENVI uses two modeling scenarios for the image-ground relationship when computing RPCs: a frame central projection and a line-central projection. Frame camera and digital (frame central) aerial photography use the frame central projection. This projection has one perspective center (S) defined as the optical center of a camera lens through which all light rays pass and create imaging signatures in the image focal plane. Any object point (A), its image point (a), and its perspective center (S) are collinear in space. The principal point (o) is defined as the foot of the perpendicular from the perspective center to the plane of an aerial photograph. Imagery from pushbroom sensors and aerial photography from line central digital cameras use the line central projection. Each scan line has its own projection center, as the following figure shows. To compute RPCs, the angles of each axis in the object space system, along with the location of the projection center, must also be determined. These are referred to as exterior orientation parameters. ENVI can automatically calculate these parameters based on GCPs that you select, but you can also edit them or manually enter them as needed. References Wang, Zhizhuo, 1990, Principles of Photogrammetry (with Remote Sensing), Beijing: Publishing House of Surveying and Mapping. McGlone, J. C., editor, 2004, Manual of Photogrammetry, Fifth Edition, American Society for Photogrammetry and Remote Sensing. ProductENVI Version6.2 Table of Contents What's New in This Release Getting Started with ENVI Explore Data Supported Data Types Open and Browse Data Open and Display Files Open from Data Manager Open NITF and MIE4NITF Files NITF Background NITF Tagged Record Extensions Background NITF PIA TREs Background Open Scientific Data Formats Open Suomi NPP VIIRS Data LAS Format LiDAR Files Download OpenStreetMap Vectors Download Digital Elevation Model Download Web Data Open Remote Datasets Remote Connection Manager Using the Open Remote Dataset Dialog Remote Data Sources Background Image Window Views Layer Manager Raster Layers Contour Layers Portals View Metadata ENVI Header Files ENVI Image Files Play Video Files Animation Animate Using Xtreme Viewer Animate a Raster Series Animate Similar NITF Image Segments Animate MIE4NITF Series or Frames Classic LiDAR Viewer Save and Restore a Session Metadata View Metadata ENVI Header Files Display Tools Keyboard Shortcuts Placemarks Raster Color Slices Profiles and Plots Scatter Plots Publish Files to Other Applications Upload Files to ENVI Repository Upload Raster to ENVI Connect Upload Vectors to ArcGIS Portal ROIs, Vectors, Annotations Region of Interest Layers Region of Interest (ROI) Tool Region of Interest Shortcuts Regions of Interest Toolbox Tools Vector Layers Vector Toolbox Tools Classic EVF/ROI to Shapefile Filter Vectors Reproject Vectors Smooth Vectors Vector to Bounding Box Vector to Centroid Vector to Feature Count Reference Vector Records Background Vector Auxiliary Files Background Annotations Layers Spatiotemporal Analysis Working with Temporal Cubes Working with Raster Series Temporal Processing Examples Statistics Global Spatial Statistics Local Spatial Statistics View Statistics File Count Features Save and Chip Output Chip ENVI Views to PowerPoint Save NITF Files Reference NITF/NSIF Background NITF PIA TREs Background NITF Tagged Record Extensions Background Pyramids Background Remote Data Sources Background Stretch Types Background Supported Data Types Background Preprocess Data Atmospheric Correction Module QUick Atmospheric Correction (QUAC) Fast Line-of-sight Atmospheric Analysis of Hypercubes (FLAASH) Example: Multispectral Sensors and FLAASH Reference QUAC Background FLAASH Background Masks Calculate Cloud Mask Using Fmask Compute AVHRR Sea Surface Temperature Create Image Mosaics Create Seamless Mosaics Create Quick Mosaics Create Pixel-Based Mosaics Manage Raster Data Build Band Stacks Build Layer Stacks Cast Raster Data Types Convert Complex Data Convert Interleave Create Binary Rasters by Automatic Thresholds Create ENVI Meta Files Dice Rasters Export Data to IDL Variables Generate Thumbnail Image Generate Test Data Import IDL Variables Landsat MSS Aspect Landsat MSS Deskew Regrid a Raster Replace Bad Lines Resize Rasters Rotate/Flip Data Stretch Data View Byte Data Radiometric Correction Tools Atmospheric Correction Radiometric Calibration Calibrate AVHRR and TIMS Data Preprocessing Workflow Other Radiometric Correction Tools Transforms Color Transforms Decorrelation Stretch Dimensionality Expansion Image Sharpening Independent Components Analysis Minimum Noise Fraction Transform Principal Components Analysis Photographic Stretch Saturation Stretch Synthetic Color Images Tasseled Cap Vegetation Suppression Analyze Data Anomaly, Change, and Target Detection Anomaly Detection Workflow RX Anomaly Detection Tool Change Detection Analysis Change Detection Workflow SAM Target Finder with BandMax Workflow Target Detection Workflow Thematic Change Tool Thematic Change Workflow Two-Color Multiview Band Math and Spectral Indices Band Math IDL Tips for Use with Band Math Band Math User Functions Band Ratios Spectral Indices Reference Burn Indices Background Gas Indices Background Geology Indices Background Mineral Indices Background Miscellaneous Indices Background Vegetation Indices Background Classification Classification Workflow Class Layers ENVI Classification Files Classification Tools Supervised Methods Adaptive Coherence Estimator Binary Encoding Constrained Energy Minimization Mahalanobis Distance Maximum Likelihood Minimum Distance Orthogonal Subspace Projection Parallelepiped Spectral Angle Mapper Spectral Information Divergence Support Vector Machine Unsupervised Methods ISODATA K-Means Post Classification Buffer Zone Images Calculate Confusion Matrices Class Statistics Classification Aggregation Classification Smoothing Classification to Pixel ROI Classification to Polygon ROI Classification to Vector Classify from Rule Images Clump Classes Combine Classes Display ROC Curves Edit Classification Images Generate a Random Sample Majority/Minority Analysis Sieve Classes Other Tools Decision Tree Build a New Decision Tree Execute a Decision Tree Edit an Existing Decision Tree Execute an Existing Decision Tree Collect Endmember Spectra ENVI LiDAR User Interface Directories for ENVI LiDAR-Generated Products Set Preferences Coordinate System Settings Exploring Data Color the Point Cloud Color the Display Color by Height, Intensity, or RGB Apply a Height Color Palette Display Height Legend Show Point Features Annotate the View Calculate Viewshed Analysis Measure Distance Between Two Points Filter Points by Height Test DEM Precision Create and Process Projects Start a New Project Import Reference Data Process the Data Generate a Density Map Define Product Outputs Select an Area to Process Set Processing Parameters Export Products Launch Screen Capture in Microsoft PowerPoint Quality Assurance Review Processing Results Correct and Reprocess Data Make Manual Changes Manage QA Change Lists Feature Extraction Example-Based Rule-Based Segment Only Create Segmentation Images Point Cloud Feature Extraction Spatial, Spectral and Textural Attributes Reference Segmentation Algorithms Background Merge Algorithms Background Rule-Based Classification Background Support Vector Machine Background K Nearest Neighbor Background Principal Components Analysis Background Filters Convolution and Morphology Filters Texture Filters Adaptive Filters Frequency Filters Geographic Positioning Tools RPC Orthorectification RPC Orthorectification Using Reference Image Reference Accuracy Assessment Background GCP File Formats Generate GCPs from Reference Image Image Registration Automatic Image Registration Workflow Overlay Grid Lines Reproject Images Select Coordinate Systems Create Coordinate System Strings Build RPCs Select Input File Parameters for Digital Cameras and Pushbroom Sensors Frame Camera Digital (Frame Central) Digital (Line Central) Pushbroom Sensor Photogrammetry Module Rigorous Orthorectification RigorousOrthoWizard RigorousOrthoLayoutManager RigorousOrthoSupportedImageFormats RigorousOrthoSelectInputFiles RigorousOrthoCollectGCPs RigorousOrthoCollectEditTiePoints RigorousOrthoReorderImagesDefineCutlines RigorousOrthoOutputParameters RPC Orthorectification Using DSM Generate Point Clouds and DSM Legacy Tools Image to Map Registration Warp and Resample Correct MODIS Bow Tie Effect Convert ASCII Coordinates Convert Map Coordinates Georeference from Input Geometry Georeference ASTER Data Georeference AVHRR Data Georeference COSMO-SkyMed Data Georeference Envisat Georeference MODIS Georeference RADARSAT Georeference SPOT Data Georeference SeaWiFS Data Build ASTER Geometry Files Build AVHRR Geometry Files Build SeaWiFS Geometry Files Build SPOT Geometry Files Reference Overview of Map Information in ENVI Frame and Line Central Projections Background NITF Map Information Background Mobility Tools Forest Cover Classification Helicopter Landing Zones Topographic Breaklines Radar Tools ALOS Radar Backscatter Spectral Tools Profiles and Plots Material Identification Image Cubes Linear Spectral Unmixing Mapping Methods Continuum Removal LS-Fit (Linear Band Prediction) Matched Filtering Mixture Tuned Matched Filtering Multi Range Spectral Feature Fitting Spectral Feature Fitting N-Dimensional Visualizer Locating Endmembers in a Spectral Data Cloud Pixel Purity Index Relative Water Depth SMACC Spectral Hourglass Workflow Spectral Library Viewer Spectral Math Spectral Resampling Reference Topographic Tools 3D SurfaceView Convert Vector Topographic Maps to Raster DEMs DEM Extraction DEMExtractionWizard Select Input Files for DEM Extraction Select Ground Control Points View, Add, and Edit GCPs Collect Tie Points View, Add, and Edit Tie Points Generate Epipolar Images Set Output DEM Projection Select DEM Extraction Parameters Examine the DEM Result IndividualProcesses Build Epipolar Images Extract a DEM Individual Processes Select Stereo GCPs Select Stereo Tie Points DEM Editing Tool Epipolar 3D Cursor Tool DEM Extraction Module Stereo Pair 3D Measurement Tool Extract Topographic Features Generate Full-Resolution Contour Lines GMTED2010 Overview Rasterize Point Data Replace Bad DEM Values Topographic Breaklines Topographic Modeling Topographic Shading Interactive Viewshed Analysis Vegetation Analysis Agricultural Stress Classification Fire Fuel Classification Forest Cover Classification Forest Health Classification NDVI Vegetation Analysis Workflow Vegetation Index Calculator Vegetation Indices Broadband Greenness Narrowband Greenness Light Use Efficiency Canopy Nitrogen Dry or Senescent Carbon Leaf Pigments Canopy Water Content Reference Canopies Plant Foliage Non-Photosynthetic Vegetation Crop Science Module Workflows Calculate Crop Metrics Calculate Crop Metrics with Spectral Index Calculate Zone Metrics Calculate Zone Metrics with Spectral Index Convert Crops To Shapefile Convert Zones to Shapefile Count Crops Create Crop Location Grid Create Zones Enhance Crops Find Crop Gaps Find Developing Hotspots Find Hotspots Find Rows and Remove Outliers ENVI Repositories ENVI Servers ENVI Modeler Getting Started with the ENVI Modeler Add Tasks to Models Specify Input Datasets and Task Parameters Extract Properties and Metadata Extract Elements From Arrays Batch-Process Data Using Iterator Nodes Apply Conditional Statements Using Filter Iterator Nodes Collect Items Using Aggregator Nodes Add Breakpoints Add Comments Specify Model Output Create Metatasks From Models Validate and Run Models Generate and Run Code From Models ENVI Modeler Examples Example: Sentinel-2 NDVI Color Slice Classification Example: Using Conditional Operators with Rasters Example: Image Registration Metatask Programming ENVI API Programming ENVI API Frequently Asked Questions Getting Started Display Control Working with Rasters Raster Metadata Map Information Data Processing using ENVITask Metatasks Custom Tasks Deploy Custom Tasks Sample Task Template Style Sheets for User Interface Elements User Interface Elements IDLContainer_UI Examples Custom Task Base Classes Custom Plot Functions Workflows The Basic Structure of a Workflow Customize Workflows Tips for Creating Effective Workflows Example: Create a Workflow Report Callback Example: Perform a Step Automatically Callback Example: Restructure a Workflow Callback Example: Show and Hide Task Parameters Custom Task Example: Allow Users to Skip a Workflow Step Custom UI Class Example: Choose Random Input Rasters Custom UI Class Example: Present Multiple Output Options in One Step Processing Large Rasters Using Tile Iterators Run ENVI Analytics at the Command Line Command Line Input and Output ENVITaskEngine Examples Bash Shell Script Example Windows Batch Script Example Data Type Support for ENVITaskEngine Advanced Concepts Error Handling Event Handling Mouse Event Handlers Keyboard Event Handlers Selection Change Event Handlers Event Handler Class Messaging Toolbox Extensions Application Control Routines ENVI Function ENVI::AddCustomReader ENVI::AddExtension ENVI::CleanupTemporaryWorkspace ENVI::Close ENVI::CreateView ENVI::DeleteLocalRepository ENVI::DownloadFromRepository ENVI::EditRepositoryPackage ENVI::GetBroadcastChannel ENVI::GetLocalPackageInfo ENVI::GetTemporaryFilename ENVI::GetView ENVI::HideExtensionFiles ENVI::LogMessage ENVI::OpenAnnotation ENVI::OpenPointCloud ENVI::OpenRaster ENVI::OpenROI ENVI::OpenVector ENVI::PublishToRepository ENVI::PingRepository ENVI::QueryPointCloud ENVI::Refresh ENVI::RegisterRepository ENVI::ReportError ENVI::RestoreSession ENVI::SaveSession ENVI::Show ENVI::UnregisterRepository Messaging ENVIAbortable ENVIBroadcastChannel ENVIBroadcastChannel ENVIBroadcastChannel::Subscribe ENVIBroadcastChannel::Unsubscribe ENVIMessage ENVIMessageHandler ENVIMessageHandler::OnMessage ENVIStartMessage ENVIProgressMessage ENVIFinishMessage ENVIPreferenceItem ENVIPreferenceItem::RestoreDefaults ENVIPreferenceItem::Save ENVIPreferences ENVIPreferences::Load ENVIPreferences::RestoreDefaults ENVIPreferences::Save ENVISecureString ENVISecureString::Dehydrate ENVISecureString::Hydrate ENVI Classic ENVI_GET_CONFIGURATION_VALUES Data Management Routines ENVIAgCalculateCropMetrics ENVIAgCalculateGetisOrd ENVIAgCropCount ENVIAgCrops ENVIAgCrops::Dehydrate ENVIAgCrops::Export ENVIAgCrops::Hydrate ENVIAgCrops::Remove ENVIAgCropsToShapefile ENVIAgEnhanceCrops ENVIAgZones ENVIAgZones::Dehydrate ENVIAgZones::GetZoneProperty ENVIAgZones::Hydrate ENVIAgZones::Save ENVIAnnotationSet ENVIAnnotationSet::AddArrow ENVIAnnotationSet::AddCircle ENVIAnnotationSet::AddPolygon ENVIAnnotationSet::AddPolyline ENVIAnnotationSet::AddRectangle ENVIAnnotationSet::AddSymbol ENVIAnnotationSet::AddText ENVIAnnotationSet::Close ENVIAnnotationSet::Dehydrate ENVIAnnotationSet::Hydrate ENVIAnnotationSet::Save ENVICalculateConfusionMatrixFromRaster ENVIConfusionMatrix ENVIConfusionMatrix::Accuracy ENVIConfusionMatrix::ColumnTotals ENVIConfusionMatrix::CommissionError ENVIConfusionMatrix::F1 ENVIConfusionMatrix::KappaCoefficient ENVIConfusionMatrix::OmissionError ENVIConfusionMatrix::Precision ENVIConfusionMatrix::ProducerAccuracy ENVIConfusionMatrix::Recall ENVIConfusionMatrix::RowTotals ENVIConfusionMatrix::UserAccuracy ENVICoordSys ENVICoordSys::ConvertLonLatToLonLat ENVICoordSys::ConvertLonLatToMap ENVICoordSys::ConvertLonLatToMGRS ENVICoordSys::ConvertMapToLonLat ENVICoordSys::ConvertMapToMap ENVICoordSys::ConvertMGRSToLonLat ENVICoordSys::Dehydrate ENVICoordSys::Hydrate ENVIDataCollection ENVIDataCollection::Add ENVIDataCollection::Count ENVIDataCollection::Get ENVIDataCollection::Remove ENVIDataContainer ENVIDataContainer::AddArray ENVIDataContainer::AddScalar ENVIDataContainer::Dehydrate ENVIDataContainer::GetArray ENVIDataContainer::GetScalar ENVIDataContainer::HasArray ENVIDataContainer::HasScalar ENVIDataContainer::Hydrate ENVIDataContainer::RemoveArray ENVIDataContainer::RemoveScalar ENVIFeatureCount ENVIFeatureCount::Close ENVIFeatureCount::Dehydrate ENVIFeatureCount::Hydrate ENVIFIDToRaster ENVIGCPSet ENVIGCPSet::Add ENVIGCPSet::ApplyOffset ENVIGCPSet::Close ENVIGCPSet::Count ENVIGCPSet::Dehydrate ENVIGCPSet::Export ENVIGCPSet::Get ENVIGCPSet::Hydrate ENVIGCPSet::Remove ENVIGeoJSON ENVIGLTRasterSpatialRef ENVIGLTRasterSpatialRef::ConvertFileToFile ENVIGLTRasterSpatialRef::ConvertFileToMap ENVIGLTRasterSpatialRef::ConvertLonLatToLonLat ENVIGLTRasterSpatialRef::ConvertLonLatToMap ENVIGLTRasterSpatialRef::ConvertLonLatToMGRS ENVIGLTRasterSpatialRef::ConvertMaptoFile ENVIGLTRasterSpatialRef::ConvertMapToLonLat ENVIGLTRasterSpatialRef::ConvertMapToMap ENVIGLTRasterSpatialRef::ConvertMGRSToLonLat ENVIGLTRasterSpatialRef::Dehydrate ENVIGLTRasterSpatialRef::Hydrate ENVIGLTRasterSpatialRef::NorthIsUpAngle ENVIGridDefinition ENVIGridDefinition::ConvertGridToMap ENVIGridDefinition::ConvertMapToGrid ENVIGridDefinition::CreateGridFromCoordSys ENVIGridDefinition::Dehydrate ENVIGridDefinition::Hydrate ENVIGridDefinition::Intersection ENVIGridDefinition::Union ENVIHydratable ENVIHydratable::Dehydrate ENVIHydrate ENVIJagwireServer ENVIJagwireServer::Close ENVIJagwireServer::Query ENVIJagwireServer::QueryDataset ENVINITF ENVINITF::Dehydrate ENVINITF::GetAssociatedFiles ENVINITF::GetBandFields ENVINITF::GetBandValue ENVINITF::GetBandValues ENVINITF::GetDESData ENVINITF::GetDESUserDefinedDataFields ENVINITF::GetDESUserDefinedDataValue ENVINITF::GetDESXMLData ENVINITF::GetSecurityFields ENVINITF::GetSecurityValue ENVINITF::GetSegmentFields ENVINITF::GetSegmentValue ENVINITF::GetTextSegments ENVINITF::GetTREFields ENVINITF::GetTRENames ENVINITF::GetTREValue ENVINITF::Hydrate ENVINITF::OpenCameraSeries ENVINITF::OpenCameraSeriesFrame ENVINITF::OpenRaster ENVINITF::OpenVector ENVINITF::WriteMetadata ENVINITFCSMRasterSpatialRef ENVINITFCSMRasterSpatialRef::ConvertFileToFile ENVINITFCSMRasterSpatialRef::ConvertFileToLonLat ENVINITFCSMRasterSpatialRef::ConvertFileToMap ENVINITFCSMRasterSpatialRef::ConvertLonLatToFile ENVINITFCSMRasterSpatialRef::ConvertLonLatToLonLat ENVINITFCSMRasterSpatialRef::ConvertLonLatToMap ENVINITFCSMRasterSpatialRef::ConvertLonLatToMGRS ENVINITFCSMRasterSpatialRef::ConvertMapToFile ENVINITFCSMRasterSpatialRef::ConvertMapToLonLat ENVINITFCSMRasterSpatialRef::ConvertMapToMap ENVINITFCSMRasterSpatialRef::ConvertMGRSToLonLat ENVINITFCSMRasterSpatialRef::Dehydrate ENVINITFCSMRasterSpatialRef::Hydrate ENVINITFCSMRasterSpatialRef::NorthIsUpAngle ENVINITFCSMRasterSpatialRef::UpIsUpAngle ENVINITFMetadata ENVINITFQuerySensorModels ENVIParameterUI ENVIParameterUI::BuildUI ENVIParameterUI::GetValue ENVIParameterUI::SetValue ENVIPointCloud ENVIPointCloud::Close ENVIPointCloud::Dehydrate ENVIPointCloud::GetPointsInCircle ENVIPointCloud::GetPointsInPolygon ENVIPointCloud::GetPointsInRange ENVIPointCloud::GetPointsInRect ENVIPointCloud::GetPointsInTile ENVIPointCloud::Hydrate ENVIPointCloud::Save ENVIPointCloud::WritePoints ENVIPointCloudFilter ENVIPointCloudMetadata ENVIPointCloudProductsInfo ENVIPointCloudProductsInfo::Dehydrate ENVIPointCloudProductsInfo::Hydrate ENVIPointCloudQuery ENVIPointCloudQuery::Close ENVIPointCloudQuery::GetPointsInRange ENVIPointCloudSpatialRef ENVIPointCloudSpatialRef::ConvertLonLatToMap ENVIPointCloudSpatialRef::ConvertMapToLonLat ENVIPointCloudSpatialRef::ConvertMapToMap ENVIPointCloudSpatialRef::Dehydrate ENVIPointCloudSpatialRef::Hydrate ENVIPreferenceItem ENVIPreferenceItem::RestoreDefault ENVIPreferenceItem::Save ENVIPreferences ENVIPreferences::Load ENVIPreferences::RestoreDefaults ENVIPreferences::Save ENVIPseudoRasterSpatialRef ENVIPseudoRasterSpatialRef::ConvertFileToFile ENVIPseudoRasterSpatialRef::ConvertFileToLonLat ENVIPseudoRasterSpatialRef::ConvertFileToMap ENVIPseudoRasterSpatialRef::ConvertLonLatToFile ENVIPseudoRasterSpatialRef::ConvertLonLatToLonLat ENVIPseudoRasterSpatialRef::ConvertLonLatToMap ENVIPseudoRasterSpatialRef::ConvertLonLatToMGRS ENVIPseudoRasterSpatialRef::ConvertMapToFile ENVIPseudoRasterSpatialRef::ConvertMapToLonLat ENVIPseudoRasterSpatialRef::ConvertMapToMap ENVIPseudoRasterSpatialRef::ConvertMGRSToLonLat ENVIPseudoRasterSpatialRef::Dehydrate ENVIPseudoRasterSpatialRef::Hydrate ENVIPseudoRasterSpatialRef::NorthIsUpAngle ENVIPseudoRasterSpatialRef::ShadowsDownAngle ENVIRaster ENVIRaster::Close ENVIRaster::CreatePyramid ENVIRaster::CreateTileIterator ENVIRaster::Dehydrate ENVIRaster::Export ENVIRaster::ExportROIs ENVIRaster::GetData ENVIRaster::Hydrate ENVIRaster::Save ENVIRaster::SetData ENVIRaster::SetTile ENVIRaster::WriteMetadata ENVIRasterIterator ENVIRasterIterator::GetData ENVIRasterIterator::Next ENVIRasterIterator::Previous ENVIRasterIterator::Reset ENVIRasterMetadata ENVIRasterMetadata::AddItem ENVIRasterMetadata::Dehydrate ENVIRasterMetadata::HasTag ENVIRasterMetadata::Hydrate ENVIRasterMetadata::RemoveItem ENVIRasterMetadata::UpdateItem ENVIRasterSeries ENVIRasterSeries::Close ENVIRasterSeries::Dehydrate ENVIRasterSeries::First ENVIRasterSeries::GetData ENVIRasterSeries::GetKey ENVIRasterSeries::Hydrate ENVIRasterSeries::Last ENVIRasterSeries::Next ENVIRasterSeries::Previous ENVIRasterSeries::Query ENVIRasterSeries::Raster ENVIRasterSeries::Set ENVIRasterStatistics ENVIRasterToFID ENVIROI ENVIROI::AddGeometry ENVIROI::AddPixels ENVIROI::AddThreshold ENVIROI::AddVectorRecords ENVIROI::ClearGeometry ENVIROI::ClearPixels ENVIROI::ClearThresholds ENVIROI::ClearVectorRecords ENVIROI::Close ENVIROI::Dehydrate ENVIROI::GetExtent ENVIROI::GrowRegion ENVIROI::Hydrate ENVIROI::PixelAddresses ENVIROI::Pixelate ENVIROI::PixelCount ENVIROI::Revert ENVIROI::Save ENVIRPCRasterSpatialRef ENVIRPCRasterSpatialRef::ConvertFileToFile ENVIRPCRasterSpatialRef::ConvertFileToLonLat ENVIRPCRasterSpatialRef::ConvertFileToMap ENVIRPCRasterSpatialRef::ConvertLonLatToFile ENVIRPCRasterSpatialRef::ConvertLonLatToLonLat ENVIRPCRasterSpatialRef::ConvertLonLatToMap ENVIRPCRasterSpatialRef::ConvertLonLatToMGRS ENVIRPCRasterSpatialRef::ConvertMapToFile ENVIRPCRasterSpatialRef::ConvertMapToLonLat ENVIRPCRasterSpatialRef::ConvertMapToMap ENVIRPCRasterSpatialRef::ConvertMGRSToLonLat ENVIRPCRasterSpatialRef::Dehydrate ENVIRPCRasterSpatialRef::Hydrate ENVIRPCRasterSpatialRef::NorthIsUpAngle ENVIRPCRasterSpatialRef::UpIsUpAngle ENVIServer ENVIServer::Dehydrate ENVIServer::GetExecutedTask ENVIServer::GetJobStatus ENVIServer::GetQueuedJobs ENVIServer::GetSubmittedTask ENVIServer::Hydrate ENVIServer::SubmitTask ENVISpectralLibrary ENVISpectralLibrary::AddSpectra ENVISpectralLibrary::Dehydrate ENVISpectralLibrary::GetSpectralSignature ENVISpectralLibrary::Hydrate ENVISpectralLibrary::RemoveSpectra ENVISpectralSignature ENVIStandardRasterSpatialRef ENVIStandardRasterSpatialRef::ConvertFileToFile ENVIStandardRasterSpatialRef::ConvertFileToLonLat ENVIStandardRasterSpatialRef::ConvertFileToMap ENVIStandardRasterSpatialRef::ConvertLonLatToFile ENVIStandardRasterSpatialRef::ConvertLonLatToLonLat ENVIStandardRasterSpatialRef::ConvertLonLatToMap ENVIStandardRasterSpatialRef::ConvertLonLatToMGRS ENVIStandardRasterSpatialRef::ConvertMapToFile ENVIStandardRasterSpatialRef::ConvertMapToLonLat ENVIStandardRasterSpatialRef::ConvertMapToMap ENVIStandardRasterSpatialRef::ConvertMGRSToLonLat ENVIStandardRasterSpatialRef::Dehydrate ENVIStandardRasterSpatialRef::Hydrate ENVIStandardRasterSpatialRef::NorthIsUpAngle ENVISpatialSubsetPointCloud ENVISpatialSubsetPointCloud::Dehydrate ENVISpatialSubsetPointCloud::Hydrate ENVIStretchParameters ENVIStretchParameters::Dehydrate ENVIStretchParameters::Hydrate ENVITiePointSet ENVITiePointSet::Add ENVITiePointSet::Count ENVITiePointSet::Dehydrate ENVITiePointSet::Export ENVITiePointSet::Get ENVITiePointSet::Hydrate ENVITiePointSet::Offset ENVITiePointSet::Remove ENVITiePointSet::Scale ENVITime ENVITime::Dehydrate ENVITime::GetString ENVITime::Hydrate ENVIURLRaster ENVIVector ENVIVector::Close ENVIVector::Dehydrate ENVIVector::Hydrate ENVIWorkflow ENVIWorkflow::Add ENVIWorkflow::Connect ENVIWorkflow::Disconnect ENVIWorkflow::GenerateMetatask ENVIWorkflow::GetStep ENVIWorkflow::IsFirstStep ENVIWorkflowStep::IsLastStep ENVIWorkflowStep ENVIWorkflowStep::GetUIObject ENVIWorkflowStep::StyleSheetHideParameters ENVIWorkflowStep::StyleSheetResetUIClass ENVIWorkflowStep::StyleSheetSetUIClass ENVIWorkflowStep::StyleSheetShowParameters ENVI Classic ENVI_BANDMAX_SELECT_BANDS ENVI_CHANGE_HEAD ENVI_COMPUTE_SUN_ANGLES ENVI_CREATE_ROI ENVI_DELETE_ROIS ENVI_EVF_CLOSE ENVI_EVF_DEFINE_ADD_RECORD ENVI_EVF_DEFINE_CLOSE ENVI_EVF_DEFINE_INIT ENVI_EVF_INFO ENVI_EVF_OPEN ENVI_EVF_READ_RECORD ENVI_EVF_TO_SHAPEFILE ENVI_FILE_MNG ENVI_FILE_TYPE ENVI_GET_FILE_IDS ENVI_GET_RGB_TRIPLETS ENVI_GET_ROI_DIMS_PTR ENVI_GET_ROI_IDS ENVI_GET_STATISTICS ENVI_OUTPUT_TO_EXTERNAL_FORMAT ENVI_PLOT_DATA ENVI_PROJ_CREATE ENVI_READ_COLS ENVI_RESAMPLE_SPECTRA ENVI_SETUP_HEAD ENVI_TRANSLATE_PROJECTION_NAME ENVI_TRANSLATE_PROJECTION_UNITS ENVI_USER_DEFINED_ANNOTATION ENVI_WRITE_COSMOSKYMED_METADATA ENVI_WRITE_DBF_FILE ENVI_WRITE_STATISTICS RGB_GET_BANDS Display Control Routines ENVIAnnotationLayer ENVIAnnotationLayer::Close ENVIAnnotationLayer::MoveDown ENVIAnnotationLayer::MoveToBottom ENVIAnnotationLayer::MoveToTop ENVIAnnotationLayer::MoveUp ENVIColorMap ENVIGridLinesLayer ENVIGridLinesLayer::Close ENVIGridLinesLayer::GetView ENVIGridLinesLayer::MoveDown ENVIGridLinesLayer::MoveToBottom ENVIGridLinesLayer::MoveToTop ENVIGridLinesLayer::MoveUp ENVIPointCloudViewer ENVIPointCloudViewer::Close ENVIPointCloudViewer::Display ENVIPointCloudViewer::DisplayShapefile ENVIPointCloudViewer::GetOpenData ENVIPointCloudViewer::GetViewExtents ENVIPointCloudViewer::SetProgress ENVIPointCloudViewer::SetViewExtents ENVIPortal ENVIPortal::Animate ENVIPortal::Close ENVIRasterLayer ENVIRasterLayer::AddROI ENVIRasterLayer::Close ENVIRasterLayer::Export ENVIRasterLayer::GetLayer ENVIRasterLayer::GetView ENVIRasterLayer::MoveDown ENVIRasterLayer::MoveToBottom ENVIRasterLayer::MoveToTop ENVIRasterLayer::MoveUp ENVIRasterSeriesLayer ENVIRasterSeriesLayer::First ENVIRasterSeriesLayer::Last ENVIRasterSeriesLayer::Next ENVIRasterSeriesLayer::Previous ENVIRasterSeriesLayer::Set ENVIROILayer ENVIROILayer::Close ENVIROILayer::GetView ENVIROILayer::MoveDown ENVIROILayer::MoveToBottom ENVIROILayer::MoveToTop ENVIROILayer::MoveUp ENVIUI ENVIUI::CreateFromDialog ENVIUI::CreateWorkflowDialog ENVIUI::Hourglass ENVIUI::LinkViews ENVIUI::PlotSpectralSignature ENVIUI::RefreshWindow ENVIUI::RunTask ENVIUI::SelectCoordinateSystem ENVIUI::SelectGridDefinition ENVIUI::SelectInputData ENVIUI::SelectPseudoRasterSpatialRef ENVIUI::SelectROI ENVIUI::SelectRPCRasterSpatialRef ENVIUI::SelectSpatialRef ENVIUI::SelectSpectralLibrary ENVIUI::SelectSpectralLibrarySignature ENVIUI::SelectSpectralSignature ENVIUI::SelectStandardRasterSpatialRef ENVIUI::SelectTaskParameters ENVIUI::ShowJobConsole ENVIVectorLayer ENVIVectorLayer::Close ENVIVectorLayer::GetView ENVIVectorLayer::MoveDown ENVIVectorLayer::MoveToBottom ENVIVectorLayer::MoveToTop ENVIVectorLayer::MoveUp ENVIView ENVIView::Animate ENVIView::Chip ENVIView::ChipToFile ENVIView::ChipToVideo ENVIView::ClearSnailTrail ENVIView::Close ENVIView::ConvertFileToWindow ENVIView::ConvertWindowToFile ENVIView::CreateGridLinesLayer ENVIView::CreateLayer ENVIView::CreatePortal ENVIView::Export ENVIView::GeoLink ENVIView::GetCenterLocation ENVIView::GetExtent ENVIView::GetLayer ENVIView::GoToLocation ENVIView::HitTest ENVIView::Pan ENVIView::PixelLink ENVIView::Reset ENVIView::Rotate ENVIView::Select ENVIView::SetExtent ENVIView::Zoom ENVI Classic AUTO_WID_MNG DISP_GET_LOCATION DISP_GOTO DISP_OUT_IMG ENVI_BATCH_STATUS_WINDOW ENVI_CENTER ENVI_CLOSE_DISPLAY ENVI_COLLECT_SPECTRA ENVI_DEFINE_MENU_BUTTON ENVI_DISP_QUERY ENVI_DISPLAY_BANDS ENVI_GET_DISPLAY_NUMBERS ENVI_GET_IMAGE ENVI_INFO_WID ENVI_REPORT_INC ENVI_REPORT_INIT ENVI_REPORT_STAT ENVI_SELECT WIDGET_AUTO_BASE WIDGET_EDIT WIDGET_GEO WIDGET_MAP WIDGET_MENU WIDGET_MULTI WIDGET_OUTF WIDGET_OUTFM WIDGET_PARAM WIDGET_PMENU WIDGET_RGB WIDGET_SLABEL WIDGET_SLIST WIDGET_SSLIDER WIDGET_STRING WIDGET_SUBSET WIDGET_TOGGLE ENVITasks List of Tasks AdditiveLeeAdaptiveFilter AdditiveMultiplicativeLeeAdaptiveFilter AgCalculateAndRasterizeCropMetrics AgCalculateAndRasterizeCropMetricsWithSpectralIndex AgCalculateAndRasterizeZoneMetrics AgCalculateAndRasterizeZoneMetricsWithSpectralIndex AgCalculateCropMetrics AgCalculateGetisOrd AgCalculateZoneMetrics AgConvertZonesToShapefile AgCountAndRasterizeCrops AgCreateAndRasterizeCropLocationGrid AgCreateZones AgCropCount AgCropsToShapefile AgDevelopingHotspotColorSlice AgEnhanceCrops AgFindAndRasterizeCropGaps AgFindDevelopingHotspots AgFindDevelopingHotspotsWithSpectralIndex AgFindRowsAndRemoveOutliers AgHotspotAnalysis AgHotspotColorSlice AgRasterizeCrops AgRasterizeZones AgSpectralHotspotAnalysis AgriculturalStressClassification ApplyGainOffset ASCIIToROI ASCIIToVector AutoChangeThresholdClassification BandMaxSubsetRaster BinaryAutomaticThresholdRaster BinaryGTThresholdRaster BinaryLTThresholdRaster BinaryMorphologicalFilter BitErrorAdaptiveFilter BufferZone BuildBandStack BuildGridDefinitionFromRaster BuildIrregularGridMetaspatialRaster BuildLayerStack BuildMetaspatialRaster BuildMosaicRaster BuildRasterPyramids BuildRasterSeries BuildTemporalCube BuildTimeSeries CalculateCloudMaskForProduct CalculateCloudMaskUsingFmask CalculateConfusionMatrixFromRaster CalculateGridDefinitionFromRasterIntersection CalculateGridDefinitionFromRasterUnion CalculateQUACGainOffset CalculateRasterThreshold CalculateRelativeWaterDepth CastRaster ChangeThresholdClassification ClassificationAggregation ClassificationClumping ClassificationSieving ClassificationSmoothing ClassificationStatistics ClassificationToPixelROI ClassificationToPolygonROI ClassificationToShapefile ColorPointCloud ColorSliceClassification ConstrainedEnergyMinimization ConvertInterleave ConvertMapToGeographicCoordinates ConvertMapToPixelCoordinates ConvertPixelToMapCoordinates CreatePointCloudSubProject CreatePointCloud CreateSubrectsFromDistance CreateSubrectsFromPixels CreateSubrectsFromROI CreateSubrectsFromTileCount CreateSubrectsFromVector DarkSubtractionCorrection DataValuesMaskRaster DecorrelationStretch DiceRasterByDistance DiceRasterByPixel DiceRasterBySubrects DiceRasterByTileCount DiceRasterByVector DimensionalityExpansionRaster DimensionalityExpansionSpectralLibrary DimensionsResampleRaster DirectionalFilter DirectionalKernel DownloadFromRepository DownloadOSMVectors DownloadSRTMRasterDEM EditRasterMetadata EmissivityFromAlphaResiduals EmissivityFromNormalization EmissivityFromReferenceChannel EmpiricalLineCalibration EndmemberCollection EnhancedFrostAdaptiveFilter EnhancedLeeAdaptiveFilter EqualizationStretchRaster ExportColorSlices ExportRasterToCADRG ExportRasterToCOG ExportRasterToENVI ExportRasterToJPEG2000 ExportRasterToKMZ ExportRasterToNITF20 ExportRasterToNITF21 ExportRasterToNSIF10 ExportRasterToPNG ExportRasterToTIFF ExportRastersToDirectory ExportVectorToKML ExportVectorToGeoJSON ExtractBandsFromRaster ExtractColumnFromArray ExtractGeoJSONFromFile ExtractRasterFromFile ExtractRastersFromRasterSeries ExtractROIsFromFile ExtractRowFromArray ExtractVectorFromFile FeatureCountToROI FilterTiePointsByFundamentalMatrix FilterTiePointsByGlobalTransform FilterTiePointsByGlobalTransformWithOrthorectification FilterTiePointsByPushbroomModel FilterVector FindRasters FindVectors FireFuelClassification FirstOrderEntropyTexture FLAASH FlatFieldCorrection ForestCoverClassification ForestHealthClassification ForwardICATransform ForwardMNFTransform ForwardPCATransform FrostAdaptiveFilter FXSegmentation GammaAdaptiveFilter GaussianHighPassFilter GaussianHighPassKernel GaussianLowPassFilter GaussianLowPassKernel GaussianStretchRaster GenerateContourLines GenerateFilename GenerateGCPsFromReferenceImage GenerateGCPsFromTiePoints GenerateIndexArray GenerateMaskFromVector GeneratePointCloudsByDenseImageMatching GenerateThumbnail GenerateTiePointsByCrossCorrelation GenerateTiePointsByCrossCorrelationWithOrthorectification GenerateTiePointsByMutualInformation GenerateTiePointsByMutualInformationWithOrthorectification GeographicSubsetRaster GeoJSONToROI GeoPackageToShapefile GetColorSliceRanges GetColorTable GetSpectrumFromLibrary GetVersion GramSchmidtPanSharpening GrayscaleMorphologicalFilter Helicopter Landing Zones HighClipRaster HighPassFilter HighPassKernel IARReflectanceCorrection ImageBandDifference ImageIntersection ImageThresholdToROI ImageToImageRegistration InverseMNFTransform ISODATAClassification KuanAdaptiveFilter LabelEntropyTexture LabelRegions LaplacianFilter LaplacianKernel LinearPercentStretchRaster LinearRangeStretchRaster LinearSpectralUnmixing LocalSigmaAdaptiveFilter LogResidualsCorrection LogStretchRaster LowClipRaster LowPassFilter LowPassKernel MahalanobisDistanceClassification MappingResampleRaster MaskRaster MatchedFilter MaximumLikelihoodClassification MedianFilter MergeROI MinimumDistanceClassification MirrorRaster MixtureTunedMatchedFilter MixtureTunedTargetConstrainedInterferenceMinimizedFilter MultiplicativeLeeAdaptiveFilter NITFToRasterSeries NNDiffusePanSharpening NormalizedEuclideanDistance OptimizedLinearStretchRaster OrthogonalSubspaceProjection ParallelPipedClassification PCPanSharpening PercentThresholdClassification PixelPurityIndex PixelScaleResampleRaster PixelStatistics PixelwiseBandMathRaster PointCloudFeatureExtraction PublishToRepository QUAC QueryAllTasks QuerySpectralIndices QuerySpectralLibrary ENVIQueryTaskCatalog ENVIQueryTask RadarBackscatter RadiometricCalibration RadiometricNormalization RankStrengthTexture RasterConvolution RasterHistogram RasterMetadataItem RasterProperties RasterStatistics RasterThresholdToVector RasterViewshed RegridRaster RegridRasterSeries RegridRasterSeriesByIndex RegridRasterSeriesByIntersection RegridRasterSeriesByUnion RegisterRasterWithGeoServer RegisterVectorWithGeoServer RemoveScanLineStriping ReprojectGLT ReprojectRaster ReprojectVector ResampleSpectrum RGBToHSIRaster RobertsFilter ROIMaskRaster ROIStatistics ROIToClassification ROIToGeoJSON ROIToKML RootStretchRaster RPCOrthorectification RPCOrthorectificationUsingDSMFromDenseImageMatching RPCOrthorectificationUsingReferenceImage RuleRasterClassification RXAnomalyDetection SAMImageDifference SetRasterMetadata SmoothVector SobelFilter SpectralAdaptiveCoherenceEstimator SpectralAdaptiveCoherenceEstimatorUsingSubspaceBackgroundStatistics SpectralAngleMapperClassification SpectralBinning SpectralIndex SpectralIndices SpectralInformationDivergenceClassification SpectralSimilarityMapperClassification SpectralSmoothing SpectralSubspaceBackgroundStatistics StringProcessing SubsetRaster TargetConstrainedInterferenceMinimizedFilter ThematicChange ThermalAtmosphericCorrection TopographicBreaklines TopographicFeatures TopographicModeling TopographicShadingUsingHLS TopographicShadingUsingHSV TopographicShadingUsingRGB TrainingClassificationStatistics TransposeRaster TwoColorMultiview UploadRasterToENVIConnect UploadVectorToArcGISPortal VectorAttributeToROIs VectorMaskRaster VectorRecordsToBoundingBox VectorRecordsToCentroid VectorRecordsToROI VectorRecordsToSeparateROI VectorToFeatureCount VegetationDelineation VegetationSuppression VideoToRasterSeries ENVITask Function ENVITask::AddParameter ENVITask::Dehydrate ENVITask::Execute ENVITask::Hydrate ENVITask::Parameter ENVITask::ParameterNames ENVITask::RemoveParameter Masking Support in ENVITasks Task Processing QueryAllTasks QueryTaskCatalog QueryTask RunTask Run Tasks from the ENVI Toolbox Run ENVITasks Asynchronously ENVIAsyncBridgeTaskJob ENVIAsyncBridgeTaskJob::OnDone ENVIAsyncSpawnTaskJob ENVIAsyncSpawnTaskJob::OnDone ENVIAbortableTaskFromProcedure ENVIAbortableTaskFromProcedure::PreExecute ENVIAbortableTaskFromProcedure::DoExecute ENVIAbortableTaskFromProcedure::PostExecute ENVITaskFromProcedure ENVITaskFromProcedure::DoExecute ENVITaskFromProcedure::PostExecute ENVITaskFromProcedure::PreExecute ENVI Classic Data Processing Routines AIRSAR_HEADER_DOIT AIRSAR_PED_HEIGHT_DOIT AIRSAR_PHASE_IMAGE_DOIT AIRSAR_POLSIG_DOIT AIRSAR_SCATTER_DOIT AIRSAR_SYNTH_DOIT ASPECT_DOIT BAD_DATA_DOIT CLASS_DOIT CLASS_RULE_DOIT CLASS_STATS_DOIT COM_CLASS_DOIT CONTINUUM_REMOVE_DOIT CONVERT_INPLACE_DOIT CROSS_TRACK_CORRECTION_DOIT DECOR_DOIT DEM_BAD_DATA_DOIT DESKEW_DOIT DESTRIPE_DOIT EMITTANCE_CALC_DOIT ENVI_AVHRR_CALIBRATE_DOIT ENVI_AVHRR_GEOMETRY_DOIT ENVI_AVHRR_WARP_DOIT ENVI_CEM_DOIT ENVI_CLOVER_DOIT ENVI_COMPUTE_SUN_ANGLES ENVI_CONVERT_LIDAR_DATA_DOIT ENVI_CUBE_3D_DOIT ENVI_DOIT ENVI_ENVISAT_GEOREF_DOIT ENVI_FILTER_DOIT ENVI_FX_EXAMPLEBASED_DOIT ENVI_FX_RULEBASED_DOIT ENVI_GEOREF_FROM_GLT_DOIT ENVI_GLT_DOIT ENVI_GRID_DOIT ENVI_ICA_INV_DOIT ENVI_NEURAL_NET_DOIT ENVI_OSP_DOIT ENVI_RADARSAT_GEOREF_DOIT ENVI_REGISTER_DOIT ENVI_SEAWIFS_GEOMETRY_DOIT ENVI_SEAWIFS_GEOREF_DOIT ENVI_SMACC_DOIT ENVI_SVM_DOIT ENVI_SYNTHETIC_COLOR_DOIT ENVI_TCIMF_DOIT ENVI_TCIMF_MF_DOIT ENVI_THERMAL_CORRECT_DOIT FFT_DOIT FFT_INV_DOIT GEN_IMAGE_DOIT HIST_EXPORT_DOIT MATH_DOIT MNF_INV_DOIT MORPH_DOIT MUNSELL_DOIT MUNSELL_INV_DOIT RADAR_INC_ANGLE_DOIT RATIO_DOIT RGB_ITRANS_DOIT RGB_TRANS_DOIT ROC_CURVE_DOIT ROTATE_DOIT RTV_DOIT SAT_STRETCH_DOIT SHARPEN_DOIT SIRC_HEADER_DOIT SIRC_MULTILOOK_DOIT SIRC_PED_HEIGHT_DOIT SIRC_PHASE_IMAGE_DOIT SIRC_POLSIG_DOIT SIRC_SYNTH_DOIT SLT2GND_DOIT SPECTRAL_FEATURE_DOIT TASCAP_DOIT TEXTURE_COOCCUR_DOIT TEXTURE_STATS_DOIT TIMS_CAL_DOIT Obsolete Routines ADAPT_FILT_DOIT CF_DOIT CLASS_CONFUSION_DOIT CLASS_CS_DOIT CLASS_MAJORITY_DOIT CONV_DOIT CONVERT_DOIT DARK_SUB_DOIT E3De E3De::AddAnnotation E3De::AddExtension E3De::Close E3De::CreateLidarFromSubrect E3De::DeleteAnnotation E3De::DisplayShapefile E3De::GenerateProducts E3De::GetOpenData E3De::GetProductionParameters E3De::GetProductsInfo E3De::GetViewExtents E3De::OpenLidar E3De::SetProgress E3De::SetViewExtents E3DLasHeader E3DLidar E3DLidar::Close E3DLidar::GetDataRange E3DLidar::GetLasHeader E3DLidar::GetPointsInCircle E3DLidar::GetPointsInPolygon E3DLidar::GetPointsInRange E3DLidar::GetPointsInRect E3DLidar::GetPointsInTile E3DLidar::Init E3DLidar::Save E3DLidar::WritePoints E3DLidarPointFilter E3DLidarSpatialRef E3DProductionParameters E3DProductionParameters::Save E3DProductsInfo ELINE_CAL_DOIT ENVI procedure ENVI::CreateRaster ENVI::CreateRasterMetadata ENVI::CreateRasterSpatialRef ENVI::ExportRaster ENVI::GetOpenData ENVI::GetPreference ENVIRaster::Subset ENVIRasterSpatialRefGLT ENVIRasterSpatialRefPseudo ENVIRasterSpatialRefStandard ENVIRasterSpatialRefRPC ENVISpectralLibrary::GetSpectrum ENVITaskParameter ENVITaskParameter::QueryProperty ENVITaskParameter::Validate ENVITask::Validate ENVI_ACE_DOIT ENVI_ADD_PROJECTION ENVI_ASSIGN_HEADER_VALUE ENVI_BATCH_EXIT ENVI_BATCH_INIT ENVI_BUFFER_ZONE_DOIT ENVI_CAL_DOIT ENVI_CONVERT_FILE_COORDINATES ENVI_CONVERT_FILE_MAP_PROJECTION ENVI_CONVERT_PROJECTION_COORDINATES ENVI_DEFAULT_STRETCH_CREATE ENVI_DEFINE_ROI ENVI_ENTER_DATA ENVI_FILE_QUERY ENVI_FX_SEGMENTONLY_DOIT ENVI_GET_DATA ENVI_GET_HEADER_VALUE ENVI_GET_MAP_INFO ENVI_GET_PATH ENVI_GET_PROJECTION ENVI_GET_ROI ENVI_GET_ROI_DATA ENVI_GET_ROI_INFORMATION ENVI_GET_SLICE ENVI_GET_TILE ENVI_GS_SHARPEN_DOIT ENVI_ICA_DOIT ENVI_INIT_TILE ENVI_IO_ERROR ENVI_LAYER_STACKING_DOIT ENVI_MAP_INFO_CREATE ENVI_MASK_APPLY_DOIT ENVI_OPEN_DATA_FILE ENVI_OPEN_FILE ENVI_PC_SHARPEN_DOIT ENVI_PICKFILE ENVI_QUAC_DOIT ENVI_QUERY_VERSION ENVI_REPORT_ERROR ENVI_RESTORE_ROIS ROI_THRESH_DOIT ENVI_ROI_TO_IMAGE_DOIT ENVI_RXD_DOIT ENVI_SAVE_ROIS ENVI_SEGMENT_DOIT ENVI_SELECT ENVI_SENSOR_TYPE ENVI_SET_INHERITANCE ENVI_SPECTRAL_RESAMPLING_DOIT ENVI_STATS_DOIT ENVI_SUBSPACE_BACKGROUND_STATS_DOIT ENVI_SUM_DATA_DOIT ENVI_TILE_DONE ENVI_TOGGLE_CATCH ENVI_VEG_INDEX_AVAILABLE_INDICES ENVI_VEG_INDEX_DOIT ENVI_VEG_SUPPRESS_DOIT ENVI_WRITE_ENVI_FILE ENVI_WRITE_FILE_HEADER GAINOFF_DOIT HANDLE_VALUE MAGIC_MEM_CHECK MATCH_FILTER_DOIT MATCH_FILTER_MT_DOIT MNF_DOIT MOSAIC_DOIT NDVI_DOIT PC_ROTATE PPI_DOIT RESIZE_DOIT ROI_THRESH_DOIT STRETCH_DOIT TMCAL_DOIT TOPO_DOIT TOPO_FEATURE_DOIT UNMIX_DOIT VAX_IEEE_DOIT Virtual Rasters ENVIBandMaxSubsetRaster ENVIBandMaxSubsetRaster::Dehydrate ENVIBandMaxSubsetRaster::Hydrate ENVIBinaryGTThresholdRaster ENVIBinaryGTThresholdRaster::Dehydrate ENVIBinaryGTThresholdRaster::Hydrate ENVIBinaryLTThresholdRaster ENVIBinaryLTThresholdRaster::Dehydrate ENVIBinaryLTThresholdRaster::Hydrate ENVICalibrateRaster ENVICalibrateRaster::Dehydrate ENVICalibrateRaster::Hydrate ENVICastRaster ENVICastRaster::Dehydrate ENVICastRaster::Hydrate ENVIDataValuesMaskRaster ENVIDataValuesMaskRaster::Dehydrate ENVIDataValuesMaskRaster::Hydrate ENVIDimensionalityExpansionRaster ENVIDimensionalityExpansionRaster::Dehydrate ENVIDimensionalityExpansionRaster::Hydrate ENVIEqualizationStretchRaster ENVIEqualizationStretchRaster::Dehydrate ENVIEqualizationStretchRaster::Hydrate ENVIFirstOrderEntropyTextureRaster ENVIFirstOrderEntropyTextureRaster::Dehydrate ENVIFirstOrderEntropyTextureRaster::Hydrate ENVIGainOffsetRaster ENVIGainOffsetRaster::Dehydrate ENVIGainOffsetRaster::Hydrate ENVIGainOffsetWithThresholdRaster ENVIGainOffsetWithThresholdRaster::Dehydrate ENVIGainOffsetWithThresholdRaster::Hydrate ENVIGaussianStretchRaster ENVIGaussianStretchRaster::Dehydrate ENVIGaussianStretchRaster::Hydrate ENVIHighClipRaster ENVIHighClipRaster::Dehydrate ENVIHighClipRaster::Hydrate ENVIIrregularGridMetaspatialRaster ENVIIrregularGridMetaspatialRaster::Dehydrate ENVIIrregularGridMetaspatialRaster::Hydrate ENVILabelEntropyTextureRaster ENVILabelEntropyTextureRaster::Dehydrate ENVILabelEntropyTextureRaster::Hydrate ENVILayerStackRaster ENVILayerStackRaster::Dehydrate ENVILayerStackRaster::Hydrate ENVILinearPercentStretchRaster ENVILinearPercentStretchRaster::Dehydrate ENVILinearPercentStretchRaster::Hydrate ENVILinearRangeStretchRaster ENVILinearRangeStretchRaster::Dehydrate ENVILinearRangeStretchRaster::Hydrate ENVILogStretchRaster ENVILogStretchRaster::Dehydrate ENVILogStretchRaster::Hydrate ENVILowClipRaster ENVILowClipRaster::Dehydrate ENVILowClipRaster::Hydrate ENVIMaskRaster ENVIMaskRaster::Dehydrate ENVIMaskRaster::Hydrate ENVIMetaspatialRaster ENVIMetaspatialRaster::Dehydrate ENVIMetaspatialRaster::Hydrate ENVIMetaspectralRaster ENVIMetaspectralRaster::Dehydrate ENVIMetaspectralRaster::Hydrate ENVIMirrorRaster ENVIMirrorRaster::Dehydrate ENVIMirrorRaster::Hydrate ENVIMosaicRaster ENVIMosaicRaster::Dehydrate ENVIMosaicRaster::Hydrate ENVIMosaicRaster::SaveSeamPolygons ENVINNDiffusePanSharpeningRaster ENVINNDiffusePanSharpeningRaster::Dehydrate ENVINNDiffusePanSharpeningRaster::Hydrate ENVIOptimizedLinearStretchRaster ENVIOptimizedLinearStretchRaster::Dehydrate ENVIOptimizedLinearStretchRaster::Hydrate ENVIPixelwiseBandMathRaster ENVIPixelwiseBandMathRaster::Dehydrate ENVIPixelwiseBandMathRaster::Hydrate ENVIQUACRaster ENVIQUACRaster::Dehydrate ENVIQUACRaster::Hydrate ENVIRadarBackscatterRaster ENVIRadarBackscatterRaster::Dehydrate ENVIRadarBackscatterRaster::Hydrate ENVIRankStrengthTextureRaster ENVIRankStrengthTextureRaster::Dehydrate ENVIRankStrengthTextureRaster::Hydrate 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https://www.researchgate.net/publication/329876745_Beta_b-Oxidation_of_Fatty_Acid_and_its_associated_Disorders
Published Time: 2018-12-18 (PDF) Beta (β)-Oxidation of Fatty Acid and its associated Disorders Article PDF Available Beta (β)-Oxidation of Fatty Acid and its associated Disorders December 2018 Authors: Satyam Prakash Janaki Medical College Download full-text PDFRead full-text Download full-text PDF Read full-text Download citation Copy link Link copied Read full-textDownload citation Copy link Link copied Citations (10)References (67)Figures (5) Abstract and Figures The lipids of metabolic significance in the mammalian organisms include triacylglycerols, phospholipids and steroids, together with products of their metabolism such as long-chain fatty acids, glycerol and ketone bodies. The fatty acids which are present in the triacylglycerols in the reduced form are the most abundant source of energy and provide energy twice as much as carbohydrates and proteins. Fatty acids represent an important source of energy in periods of catabolic stress related to increased muscular activity, fasting or febrile illness, where as much as 80% of the energy for the heart, skeletal muscles and liver could be derived from them. The prime pathway for the degradation of fatty acids is mitochondrial fatty acid β-oxidation (FAO). The relationship of fat oxidation with the utilization of carbohydrate as a source of energy is complex and depends upon tissue, nutritional state, exercise, development and a variety of other influences such as infection and other pathological states. Inherited defects for most of the FAO enzymes have been identified and characterized in early infancy as acute life-threatening episodes of hypoketotic, hypoglycemic coma induced by fasting or febrile illness. Therefore, this review briefly highlights mitochondrial β-oxidation of fatty acids and associated disorders with clinical manifestations. Carnitine shuttle system … : Energetics of Palmitic acid Oxidation … Stages of mitochondrial β-oxidation … : Inherited disorders of beta oxidation defect … Reactions occurring during β-oxidation steps of fatty acid … Figures - uploaded by Satyam Prakash Author content All figure content in this area was uploaded by Satyam Prakash Content may be subject to copyright. Discover the world's research 25+ million members 160+ million publication pages 2.3+ billion citations Join for free Public Full-text 1 Content uploaded by Satyam Prakash Author content All content in this area was uploaded by Satyam Prakash on Dec 23, 2018 Content may be subject to copyright. Review Articl e Beta (β)-Oxidatio n of Fatt y Acid and i ts associated Disorders Satyam Prakash Assistant Professor, Dept. of Biochemistry, Janaki Medical C ollege Teaching Hospital, Janakpu r, Nepal Mobile: +977-9841 603704, E-mail: sprakashy2424@gm ail.com Accepted 18 Decemb er, 2018 The lipids of metabolic significance in the m ammalian organisms inc lude tria cylglycer ols, phospholipids and ster oids, togethe r w ith products of their m etabolism such as long-chain fatty acids, glycerol and k etone bodies. The fatty acids which a re present in the triac ylglycerols i n the reduced form are the m ost abundant source of energy and prov ide energy t wice as much as carbohydrates and proteins. Fatty acids repre sent an important sou rce of ene rgy in periods of catabolic stre ss re lated to increased m uscular activity, fasting or febrile illness, where as much as 80% of the energy for the heart, skeletal muscles and liver could be derived from them. The prime pathway for the degrada tion of fatty acids is m itochondrial fatty a cid β-oxidation (FAO). T he relationship of fat oxida tion w ith the utili zation of carbohydrate as a source of ene rgy i s com plex a nd depends upon tissue, nutritional state, ex ercise, development and a variety of other influence s such as infection and other pathologica l states. Inherited defects f or most of the FAO enzym es ha ve been identified an d c haracteriz ed in ea rly infancy as acute life-threatening episode s of hypoketotic, hy poglycemic coma induced by fasting or febrile illness. Therefore, this review briefly highlights m itochondrial β-oxidation of fatty a cids and associated disorders with c linical manifestations. Keywords: Carnitine, Fatt y acid β-oxidation, Jam aican vomiting sickness,Stoichiometry, Sudden in fant death syndrome. INTRODUCTION Mitochondrial β-oxidati on of fatty acids plays an imp ortant role in energy production, especially during star vation, prolonged fasting or low intensity exercise. The princip al sources of fatt y acids for o xidation are di etary and mobilization of triacylglycerols mainly stored in adipocytes of adip ose tissue (Lopaschuk et al., 1994; McGarry & Foster, 1980). The release of metabolic energy, in the form of fatt y acids, is c ontrolled by a complex s eries of interrelated cascades that result in the activation of hormone-s ensitive lip ase, which h ydrolyzes fatty acids from triacylglycerols and diacylgl ycerols (Gibbons et al., 2000). The f inal fatty acid is released from monoacylglycerols through the action of monoacylglycerol lip ase, an enzyme active in the absence of hormonal stimulation. Once released, these fatty acids travel through t he b lood t o other tissu es such as muscle where they are o xidized to provide energy through the mitochondrial β-oxidation pathway. Mitochondria, as well as peroxisomes harbor all enzymes necessary f or FAO. Mitochondria are the m ain site for t he oxidation of plasma free fatty acids or lipoprotein associated triglycerides. The u se of fatty acids by the liver provides energy f or gluconeogenesis and ureagenesis (Liang et al; 2001). Equally important, the liver uses fatty acids to s ynthesize ketones, which serve as a fat derived fuel for the brain, and thus further reduce the need for glucose utilization. More than a dozen genetic defects in t he fatty acid oxidation pathway a re currently known. Nearly all of these defects present in early infancy as acute life-threatening episodes of hypoketotic, hypoglycemic com a induced b y fasting or febrile illness (Robert MO et al., 2009). Recognition o f the Vol. 5 (1), pp. 1 58-172, Decem ber, 2018 ©Global Science Rese arch Jour nals International Journal of Clinica l Biochemistry Author(s) retain the c opyright of this article. scienceresearchj ournals.org/ Int'l J. Clin. Biochem. 159 fatty acid oxidation disorders is o ften difficult b ecause patients can appear well until exposed to prolong ed fasting, and screening tests of met abolites m ay not always b e d iagnostic. T herefor e, th is review briefly highlights the overall pathway of mitochondri al β- oxidation of fatty acids and its associated deficien cies with its clinical correlation. Historical Preview of β- oxidation Fatty acids are a key source of energ y in animals. George Franz Kno op, a Germ an biochemist in 1904 studied the biological degradati on of fatty acid with his classical experiments which l ed him to formulate the theory of β-oxidation (Knoop, 19 04). His experiments used fatty ac ids with phenyl residu es in place of th e terminal meth yl g roups. The phenyl r esidue was not metabolized which served as a r eporter gr oup and was excreted in the urine. During his ex periment, Knoo p fed phenyl substitu ted fatty acids with an odd number of carbon atoms, like p henylpropionic acid (C 6 H 5-CH 2-CH 2- COOH) or phen ylvaleric acid (C 6 H 5-CH 2-CH 2-CH 2-CH 2- COOH) to dogs, and isolated hipp uric acid (C 6 H 5-CO-NH- CH 2-COOH), the conjugate of b enzoic acid and gly cine from their urine. In contrast, the excretory products in urine were phenyl-substituted fatty acids with an even number of carbon atoms, such as phenylbutyric acid (C 6 H 5-CH 2-CH 2-CH 2-COOH), were degraded to phenylacetic acid (C 6 H 5-CH 2-COOH) and excreted as phenylaceturic acid (C 6 H 5-CH 2-CO-NH-CH 2-COOH). These annotations led K noop to propose that the oxidation of fatty acids begins at c arbon atom 3, t he β- carbo n, and th at the resulting β -ke to acids are cleaved between th e α-carbon and β -carb on to yield fatt y acids shortened b y two carb on atoms. Knoop's e xperiment on biological degr adation incit ed th e idea th at fatt y acids are degraded in a stepwise method by success ive β- oxidation. Henry Drysdale Dakin follo wed Knoop's preliminar y study and executed analogous experiments with phenylpropionic acid (Dakin, 1909). He isolated t he glycine conjugates of the fol lowing β-o xidation intermediates: phenylacr ylic acid (C 6 H 5-CH=CH-COO H), β-phenyl-β-hydroxypropionic acid (C 6 H 5-CHOH-CH 2- COOH), and benzoyl acetic acid (C 6 H 5-CO-CH 2-COOH) next to h ippuric acid. At the same tim e, t he unsub stitute d fatty acids are d egraded by β-oxidation and converted to ketone bodies in perfused livers which were verified by Embden and coworkers. As a result, b y 1 910 the crucial information needed for formulating the pathway of β - oxidation was available. Due to consistent effort of researchers, a fter a 30-year period of l ittle progr ess the oxidati on of fatty acids in c ell- free p reparations from liver was demonstrat ed by Munoz and Leloir in 1 943, and Lehninger in 1944. Their endeavor came true with t he st age for the complet e elucidation of β-oxid ation. The studies and d etailed investigations L ehninger with ce ll-free s ystems c onfirmed the need for energy as ATP to spark the oxidation of fatty acids and to be essential for the activation of fatty acids. Wakil and M ahler, as well as b y Kornberg and Pricer revealed acti vated fatty acids to be th ioesters formed from fatty acids and co enzyme A. This pr ogress was only made pr omising by prio r studies of Lipmann and his collaborators who isola ted and distinguished coenzyme A. T he structu re of active acetate is acet yl-CoA was proved by Lynen and co-workers. They also determined that the acetyl-CoA was identical with the two-carbon fragment removed from fatty acids during t heir degradation (L ynen, 1952-1953). Finally, t he sub-cellular location of the β-oxidation system was est ablished by Kennedy and Lehninger,who confirmed that mitochondria were the cel lular c omponents which are most active during fatt y acid oxidation. The mito chondrial site of this path way agr eed with the observed coup ling of fatty acid o xidation to the citr ic acid c ycle and to oxid ative phosphorylation. Moreover, in 1950s , the laboratories of G reen in Wisconsin, Lynen in Munich, and Ochoa in New York demonstrated the dir ect evidence for th e proposed β- oxidation cycle by enzyme studies which were greatly facilitated by recently develop ed techniques of protein purification and by the use of sp ectrophotometric enzyme assays wi th chemically synthesized inter media t es of β- oxidation as substr ates (Vance &Vance, 2002). Although, several studies were carried out to confirm the steps mitochondrial β –oxidation, but the initial and conclusive remarks b y Franz Kno op on β–o xidation is still considered as remarkable disc overy in bioch emistry. Hence, β–o xidation is also known as Kn oop‘s path way or β –oxidation Knoop‘s p athway. β-oxidation of fatty aci d β-oxidation o f fatty acid is defined as a m etabolic pathway that oxid izes fatty ac ids, and gen erates fatty acyl-CoA ( a thioester of fatt y acid and CoA) and acetyl CoA which consists of a series of four rep eated reactions, in which a molecule of acet yl CoA is generated, and an end product of the fatty acid by be ta-oxidation is also acetyl C oA. Since, oxidation at the β-position of the fatt y acyl-CoA was performed step wise, it wa s n amed β- oxidation. Fatt y acids are oxidized by most of t he tissu es in the body. However, brain, erythrocytes and adrenal medulla cannot utilize fatty acids for energy requirement (Gurr &Harwood , 19 91; Schulz,1 985; Schulz & Kunau,1987). Th e four steps a re in volved in β-oxidation spiral of fatty acid m etabolism which is oxidation, hydration, a second oxidation, and finall y thiolysis. These happens in repeating cycles th rough the sequential removal of 2 carbons and p roduction of acetyl-CoA, which th en enters the Krebs cycle for oxidation and A TP production. Another t arget of acetyl-C oA is the pr oduction of ketone b odies in the li ver that are elated to tissues like the heart and b rain for release of energy dur ing starvation. Fatty acids with an odd numb er of carbons in Prakash 160 the acyl chain are left at the end with propionyl-CoA, which is converted to succinyl-CoA t hat then also enters the Kr ebs c ycle. Furthermo re, un saturated fatty acids with bonds in t he cis con figuration require thr ee s eparate enzymatic steps to prepare th emselves for th e β- oxidation p athway (G ervois et al., 2 000; Thorpe & Kim,1995; Hiltunen & Qin 2000; W anders,2001). Activation of fatty aci d Cytosolic Fatty Acid Activation: The transport of fatt y acids between organs occurs either in the form of tri acylglycerols associated with lip oproteins or as un esterified fatt y acids c omplexed to serum albumin. The h ydrolysis of triac ylglycerols occurs outside of cells b y lipoprotein lipas e to yield free fatt y acids. Even though a number of stu dies has been carried out w ith isolated cells from heart, liver, and adipose tissue b ut t he mechanism by which free fatty acids enter cells r emains poorly implicated (Kunau et al., 1 995). Numerous assumed fatty acid transport proteins have been identified (Kunauet al., 1995; Ab umrad N et a l., 1999). However, their specific function(s) in fatty acid uptake and their molecular mechanisms remain to be clarified. For oxidation, fatty acids w ith carbon chains more than 14 carbons need a ctivation before passing through the mitochondrial memb rane. Fr ee fatt y acids obt ained from diet or which are stored in the adip ocytes are ma inly 14 carbons or more in length. Fatty ac ids having ≤12 carbons can surpass activati on and can easily pass through the m itochondrial membrane. Fatt y acids with long chain once cross t he p lasma m embrane either diffuse or are transported to mit ochondria, p eroxisomes, and the endopl asmic reticulum w here the y are acti vated by conversion to their CoA thio esters. The mechanism of t ransfer o f fatt y acids b etween membranes i s a facilitated pr ocess or occurs b y s imple diffusion is an unres olved issu e. The id entification of low- molecular-weight (14-15 kD a) fatty acid bin ding proteins (FABPs) in t he cytosol of various anim al t issues prompted the sugg estion that these pr oteins may function as carriers of fatty acids in the cytosolic compartment (Coe &Bernlohr, 1998). FABPs may also be involved in the cellular up take of fatty acids, their intr acellular storage, or the delivery of fatty acids to sites o f their utilization. The metabolism of fa tty acids requires their prior activation b y conversion to fatt y acyl-CoA th ioesters. The activating enzymes are ATP-dependent acyl-CoA synthetases/thiokinase, which catalyze the for mation of acyl-CoA. Fatty ac yl-CoA is formed by the formation of thioester bond between the carbo xyl g roup of the fatty acid and the thiol group of coenzyme A. This reaction also involves use of energy, by breakdown of A TP to AMP + P P i and which is a n irreversible reaction. Fatty acyl-CoA formed in the c ytosol can also be used for synthesis of phospholipids and triacylglycerols. Transport of fatty acy l CoA to mitochondria: Carnitine Shuttle System and Transport Mec hanism Since, the mitochondrial m embrane is impermeable to acyl-CoAs, the organ liver and oth er tissu e mitochond ria is unable to oxidize fatty acid s o r fatty acyl-CoA‘s . Ac yl- CoAs use the carnitine shutt le to be imported into mitochondria. Fatty a cyl-CoA thioesters th at are formed at th e outer mitochondrial membrane cannot dir ectly enter the mit ochondrial matri x, where the enzymes of β- oxidation are located, because the inner mitochond rial membrane is impermeable to CoA and its derivatives. The rever sible tr ansfer o f fatt y acyl residues from CoA to carnitine is catalyzed by carnitin e pa lmitoyltransferase I (CPT I), which is an enz yme o f the outer mitochondrial membrane. There are two isoforms t hat are imp ortant for FAO. C PT1A (ge ne CPT1A), also called liver CPT1, is not only expressed in t he liver, but also in the brain, kidney, lung, spleen, intestin e, pancreas, ovar y and fibroblasts. CP T1B (gene CPT1B) is the m uscle isoform that is highly expressed in heart, skeletal muscle and testis. Both proteins are present at th e outer mitochondrial membrane and are sensitive to inhib ition by malonyl-CoA. Carnitin e acylcarnitine t ranslocase (CAC T, SLC25A20) exchanges acylcarnitines for a free carnitine molecule f rom the insid e. O nce the a cylcarnitines have entered the m itochondria, CPT2 (gene CPT2), located at the mitochondrial in ner membrane, reconverts the acylcarnitines into thei r CoA esters, which can then undergo FAO (Ramsay et al., 2001; Bonnefont et al., 2004). Int'l J. Clin. Biochem. 161 Fig 1: Carnit ine shut tle system CPT1C is a brain-spec ific CP T with a currentl y un known function (Price, 2002). B oth CPT1 and CP T2 are primarily involved in th e import of (dietary) long-ch ain acyl-CoAs, such as palm itoyl-CoA, oleo yl-CoA, and lin oleoyl-CoA. Alternatively, carnitine can be con verted in the mitochondrial mat rix int o an acylcarnitine by th e action of CPT2 or carnitine acetyl t ransferase (CAT, gene CRA T). These a cylcarnitines can cross t he mito chondrial membrane also in t he opposit e dir ection via CAC T, resulting in the transport of these a cylcarnitines into the cytosol. Acylcarnitin es can als o cross th e plasma membrane, but th e mechanism is currently u nknown. After cr ossing th e plasm a memb rane, acylca rnitines are excreted from th e bod y via either urine or bile. This detoxification m echanism is especially im portant when acyl-CoAs accumu late; for example, in disorders of mitochondrial FAO. The fatty acyl-CoA present in th e mitochondrial matri x is ready f or β-oxidation by the enzymes p resent there to yield acetyl-CoA which then enters the Kreb‘s cycl e to give energy. This carnitine sh uttle is a rate limiting step in the oxidation of fatty acids in the mito chondria and thus fatty acid oxidation can b e regulated at this step. Malon yl CoA, a n intermediate of fatty acid sy nthesis present in the cy tosol is a n inhibitor of carnitine acyltransferase I. This indicat es that when fatt y acid synthesis is in progress, oxidation of fatty acid cannot occur at the same time as the carnitine shuttle is imp aired by inhibition of carnitine acyltransferase I. Mitochondrial oxidation of f atty acids takes place in 3 stages: First Stage : β- oxidation pa thway In th is stage, th e fatty acids undergo oxidative removal of successive two-carbon units in the form of acetyl-CoA, starting from t he carb oxyl end of the fatty acyl chain. For example, the C-16 fatty acid palmitic acid (palmitate at pH 7) undergoes 7 pass es through this oxidative sequence, in each pass losing two carbons as acetyl- CoA. A t the end o f s even c ycles, the last t wo c arbons of palmitate (originally C-15 a nd C-16) are left as acetyl- CoA. The overall result is the conversion of 16-carbon chain of palmitate to 8 two-c arbon acetyl-CoA molecul es. Second Stage: Citric acid cycle In th is stage of fatty acid oxidation, the acetyl residues of acetyl-CoA a re o xidized to CO 2 via the citric aci d cycle, which also takes pl ace in the mit ochondrial matrix. Acetyl-CoA d erived f rom fatty acid oxidation, t hus, ente rs a final c ommon pathway of oxidation along with acetyl- CoA derived from glucose via glycol ysis and pyruvate oxidation. Third Stage: M itochondrial respiratory Chain The first two stages of fatty acid oxidation pro duce the electron carriers, NADH and FADH2, which in the third stage donate electrons to the mitochondrial respiratory chain,through which electrons a re carried to oxygen. Coupled to th is flow of electrons is the phosph orylation of ADP to ATP. Thus, energy released by fatty acid oxidation is conserved as ATP. Prakash 162 Fig 2: Stages of mitochon drial β-oxid ation Mitochondrial β-oxida tion The enzymes of β-oxidation either are associated w ith the inn er m itochondrial membrane or are located in the mitochondrial matri x. Four acyl-CoA dehydrogenases with different but o verlapping chain length specificities cooperate t o assure the complete degradation of all fatt y acids that can b e m etabolized b y mitochondri al β- oxidation. The nam es of the four dehydrogenases, short- chain, medium-chain, long-chain, and very-long-chain acyl-CoA deh ydrogenases, reflect t heir ch ain-length specificities. Different steps The first stage of fatty acid o xidation for the simple cas e of a saturated chain with an even numb er of carbons, and for the sli ghtly more complicated cases of unsaturated and odd-num ber chains, w ill now be described in det ail. Once the fatty acids are transported t o t he m itochondrial matrix via carnitin e p athway, β-oxidation of fatty acyl- CoA (n carbons) occurs within the mitochondria in four steps. Each cycle o f B-oxidation, liber ating a two carbon unit-acetyl CoA, occurs i s a sequence of f our reactions: Int'l J. Clin. Biochem. 163 Fig 3: Reaction s occurring du ring β-oxidation steps of fatty acid Oxidation Fatty acyl-CoA is act ed upon by an enzym e acyl-CoA dehydrogenas e which is FAD dependent enzyme. Fatt y acyl-CoA undergoes deh ydrogenation and forms a trans- double bond at the α and β ca rb ons to form trans- Δ 2-enoyl-CoA. Acyl-CoA d ehydrogenase are present as three isoenzymes each specific for a particular carbo n chain length (short, intermediate and long). The electr ons which were removed from the fatty acyl-CoA chain are transferred to FAD w hich gets reduced to FADH 2. This FADH 2 immediately via the El ectron Transport System gets converted to ATP molecul es. Hydration Enoyl-CoA hydratase o r cronotase catalyzes this reaction where one molecul e of water i s added trans-Δ 2-enoyl- CoA. Hydration occurs at the double bond resulting in the formation of β-hydroxyacyl-CoA, also c alled as 3- hydroxyacyl-CoA. Oxidation β-hydroxyacyl-CoA un dergoes dehydrogenation to form β-ketoacyl-CoA in t he presence of β-hydroxyacyl- CoA dehydrogenase. The electrons available as a result of dehydrogen ation a re accepted by NA D+ to f orm NADH H+which imm ediately e xchanges th ese electrons with oxygen in the Electron Transport System to form ATP molecules. Thiolysis or Thiocl astic scission This reaction is call ed as thiolysis as acyl-CoA acetyltransferas e (also know n as thiol ase or β- ketothiolase) Prakash 164 In the presence of CoA-SH which causes the cleavage of β-ketoacyl-CoA to form acet yl CoA and the thioester of the original fatty acid with two carbons less. This cleavag e occurs as the β carbon ketone group is a good target for nucleophilic attack b y the thiol (-SH) group of t he c oenzyme A. The sho rtened acyl-CoA then undergoes an other cycle of o xidation, st arting with the reaction catalyzed b y acyl-CoA d ehydrogenase. Beta- ketothiolase, hydroxyacyl d ehydrogenase and eno yl- CoA hydratase a ll have b road specificit y with respect to the length of th e ac yl g roup. Thus, by r epeated tu rns of t he cycle, a fatty ac id is degraded to ac etyl-CoA molecules with one being produced every turn until t he last cycle, wherein two are produced. Acetyl CoA formed from the above steps now ent ers the K reb‘s c ycle to get ox idized to CO 2 and H 2 O.The β-oxid ation of f atty acids completes in a c yclical manner (David and Cox, 5 th ed.; Jerem y 7 th ed.; Reginald, 5t h Edition; Satyanarayana. 3rd Ed ition). Stoichiometry of β-oxidation Each β-oxidatio n cycle can be rep resented as fol lowing: C(n)Acyl-CoA + CoA-SH + FAD + NA D+ + H 2 O → C(n-2)Ac yl CoA + Acetyl CoA + FADH 2 + NADH + H+ Complete oxidation of Palm itoyl CoA can be rep resented as followi ng: (Equation A) Palmitoyl CoA + 7CoA-SH + 7 FAD + 7NA D++ 7H 2 O → 8Acetyl CoA + 7FADH 2+ 7NA DH + 7H+ Converting NADH and FADH 2 to their corresp onding ATP equivalents from the abo ve equation gives us: (Equation B) Palm itoyl CoA + 7CoA-SH + 7O 2 + 28P i + 28ADP → 8A cetyl CoA + 28ATP + 7H 2 O And af ter Acet yl C oA molecules enter Kre b‘s cy cle and Electron Transport S ytem, resulting ATP is sho wn below: 8Acetyl CoA + 16O 2 + 80Pi+ 80ADP →8CoA + 80ATP + 16CO 2 16H 2 O Thus complet e energy release is sho wn in the following equation by combining equation A and B: Palmitoyl CoA + 23O 2 + 108P i + 108ADP → CoA + 10 8ATP + 16C O 2 + 23H 2 O Oxidation of of palmitic acid yields 7 NADH + 7 FADH2 + 8 acetyl-CoA in 7 cycles of mitochondrial b eta oxidation. E very acetyl-CoA yields 3 NAD H + 1 FADH2 + 1 GTP (=A TP) during Krebs cycle. Cons idering an average production of 2.5 A TP/NADH and 1.5 ATP/FADH2 u sing the respiratory ch ain, 108 A TP molecules are p roduced. However, 2 A TP molecules were consumed du ring the initial activation of Palm itate to Pa lmitoyl-CoA which is o xidized in the mitochondr ia. So, net energ y outp ut = (108 - 2) = 106 A TP with ne w concept in modern practice but≈129 ATP wi th an old concept. One m olecule of NAD H gives 2.5 m olecules of ATP and one molecule of FADH 2 gives 1.5 molecules of ATP with new conc ept but 3 molecules of ATP and one molecule of FAD H 2 gives 2 m olecules of ATP in the Electron Transp ort System (ETC) h ad been outd ated and were practiced earlier. Table 1: En ergetics of Pal mit ic acid Oxidati on Mechanism ATP Yield (New Concept) ATP Yield (Old Concept) 1.β-oxidation 7 c ycle s 7 FADH 2 and 7 NA DH a re gene rated when o xidized by ETC 7FADH 2- 7 x 1.5 = 10.5 7 NADH-7 x 2.5 = 17.5 7 x 2 =14 7 x 3= 21 From 8 acet yl CoA 1 a cetyl CoA =3 NADH + 1 FADH 2 + 1 GTP Oxidized b y citric acid cycle, each acetyl CoA provides ATP 10 x 8 = 80 12 x 8 =96 Total Energ y from one mole o f palmitoyl CoA 108 131 Energy utilize d for activation (Formation o f Palmitoyl CoA) 2 2 Net yield of oxida tion of one mole of palmitate 108-2 =106 131-2 = 129 Regulation of Fatty Acid Oxidation Since mitochondrial β-o xidation functions either directl y to produce A TP, or to produce ketone bodies for A TP generation by p eripheral tissues, the rate of β-oxidation flux is in tegrated wi th the oxidation of oth er su bstrates, particularly glucose. T his is achieved by c ontrol bo th a t the level of entry of fatty a cids into the m itochondrion, and by further intra-mitoch ondrial c ontrols. A m ajor control on b-oxidation, and a crossover b etween fatty acid metabolism and carbohydrate oxidation, was elucidated by McGarry & Foste r in the 197 0s (McGarry & Int'l J. Clin. Biochem. 165 Foster,1980). When carbohydrate is plentiful, its mitochondrial oxid ation caus es accum ulation of cit rate within the mitochondrion which ma y then be exported. The carnitine shuttle is a rate limiting step in the oxidation o f fatty acids in the mitoch ondria and thus fatty acid oxidation can b e regulated at this st ep. Malon yl CoA, an intermediate of fatty acid synthesis p resent in the cytosol is an inhibitor of carnitine a cyltransferas e I. This indicates th at w hen fatty a cid synthesis i s in pro gress, oxidation of f atty acid c annot occur at the sam e time as the carnitin e shut tle is impaired by inhibition of carnitine acyltransferase I. F atty acid oxidation is also r egulated by the acetyl C oA to C oA ratio:as the ratio incr eases, the CoA-requiring th iolase(the enzyme part icipating in β- oxidation) reaction decreases. W hen[NADH]/[NAD+] ratio increases, the enzyme β-h ydroxyacyl-CoA dehydrogenase is inh ibited (DiMauro &DiMauro, 1973; Doege & Stahl , 200 6; Kim &Simon , 2004). Disorders of Mitoc hondrial Beta oxidation Mitochondrial fatt y acid β-oxidation disord ers (FAODs) are a heterogeneous group of de fects about 2 0 defects in fatty acid transport and mitochondrial β-oxidation. They are inh erited as autosomal recessive disorders and have a wide range of clinical presentati ons. FAODs have a varied pres entation, with either n eonatal ons et with hyperammonemia, transient hypoglycemia, metabolic acidosis, cardiom yopathy and sudden death or late onset with neuropath y, m yopathy and retinopath y (Bruno & Dimauro, 20 08; Sh ekhawat et al., 2005). Most cases with FAODs are n ow identified using new born screening by mass spectrometr y (MS/MS) of b lood spots. Pregnancies of mothers heteroz ygous for FAOD have been associated wit h development of severe pre-eclampsia, acute fatty liver of pregnanc y and HE LLP syndrome (hemolysis, elevated liver enzymes, low p latelets) in m others and intrauterine gr owth r etardation in inf ants (Roe & Mo chel, 2006; Preece & Green , 2002). The first inherited defects in the FAO pathway w ere identified in th e 1970s, carnitine p almitoyl transferase 2 (CPT2) defici ency in 1973, primary carnitine deficienc y in 1975 and medium chain acyl-coenz yme A (CoA) dehydrogenase (MCAD) deficiency in 1976 (Karpati et al., 19 75; Gregersen et al., 197 6). Most FAO enzymes were pu rified in th e 19 80s (Furuta S et al., 1981), followed by the cloning of the individual gen es and the subsequent identification of disease-causing mutations in patients (Matsubar a et al.,1990; Yokota et al., 19 90; Kelly; 1990). For almost each enz yme involved in FAO, inherited defects have been described (Wanders et a l., 1999; Rinaldo et al., 2002; Sand er &Ronald, 2010). These inclu de glut aric aciduria type 2, primary c arnitine deficiency and deficienci es of CPT1, CACT, CP T2, VLCAD, MTP (including isolated L CHAD o r thiolase), MCAD, M/SCHAD, S CAD and 2,4-dienoyl CoA redu ctase (DECR). Interestingl y, CPT1b, crotonas e, MCKAT and DCI deficiency ha ve not been identi fied as of yet (Rina ldo et al., 200 2) In general, FAO d efects have three different presentations (Wanders et al., 19 99; Rinaldo et al., 2002). The f irst is the hepatic presentation, which is a severe, ofte n lethal, disease in infancy or th e neonatal period wi th h ypoketotic h ypoglycaemia and Reye-like syndrome. This disease is triggered by a catabolic state, for example d uring intercurrent inf ections. Most importantly, this condition can be prevented, which is the main reason for th e inclusi on o f FAO de fects in n eonatal screening programmes. During infancy, patients may also present with cardiac symptoms such as dilated or hypertrophic c ardiomyopathy and/or a rrhythmias. Alternatively, FAO defects might present as a mild er, later (‗adult‘) onset d isease. This form is characteriz ed by exercise-induced myopathy and rhabdomyolysis (Sander &Ronald, 2010). Severely affected p atients may displ ay combinations of all thre e pr esentations. In additi on, FA O defects ha ve been associated with su dden infant death, Jamaican Vomiting Sickness t hat may have b een caus ed by hypoketotic hypoglycaemia or cardiac disease. Mitochondrial fatty ac id oxidation disorders com prise 4 groups: (1) Disorders of the entry o f l ong-chain fatty acids into mitochondria, (2) I ntramitochondrial β-oxidation defects of l ong-chain fatty acids affecting m embrane bound enzymes, (3) β-oxidation defects of sh ort- and m edium- chain fatt y acids affecting enzymes of the mito-chondrial matri x and (4) Dis orders of im paired electron trans fer t o th e respiratory chain from mitochondrial β-oxidation (Blau, 2014). Classification of FAODs The diff erent f atty acid oxidation d isorders (The Philadelphia Guide, 2005) could be classified as follo ws: (1) Disorders of plasma membrane functions Carnitin e uptake defect L ong-chain fatty acid transport/bin ding defect (2) Disorders of fatty ac id transport across the mitochondrial mem branes CPT I de ficiency CAC T deficiency CPT II de ficiency\ (3) Disorders of long-chain fat ty acid β-oxi dation V LCAD deficiency Tri functional protein d eficiency and isolated long-chain L3-hydroxyl-COA dehydrogenase de ficiency (4) Disorders of medium-chain fatty aci d β-oxidation MCAD de ficiency Prakash 166 Medium- and short-chain L 3-hydroxyl-COA dehydrogenase deficienc y Medium-chain 3-ketoacyl-CoA th iolase deficiency (5)Disorders of short-chain fatty acid β-oxi dation: SCAD de ficiency Table 2:Inh erited d isorders of beta ox idation defect Defect Clinical manifes tations of de fect Hepatic Cardiac Skeletal Muscle Acu te Chronic Carnitine cycle CTD (+) CPT-1 Trans CPT-2 (+) β-Oxidation c ycle Acyl-CoA dehydrogenases VLCAD MCAD SCAD 3-Hydroxyacyl-CoA dehydrogenases LCHAD SCHAD MCKT DER CPT, carnitine-palm itoyl transferase; CTD, carnitine-trans porter defect; DER, 2,4-dienoyl-coen zyme-A reductase; LCHAD, long-chain 3-hydroxyac ly-coenz yme-A dehydroge nase, MCAD, m edium-chain acyl-coenzyme-A dehydrogen ase; MCKT, m edium-c hain ke toacyl-CoA thi olase; SCAD, s hort-chain acyl-coen zyme-A dehydrogenase; SC HAD, s hort-chain 3-hydro xyacyl-coenz yme-A d ehydrogenase; TRANS, carnitine/acylc arnitine translocase; VLCAD, very-long-c hain acyl-coenzyme-A deh ydrogenase Sudden Infant Death Syndro me(SIDS) Disorders of fatty acid oxidation p lay a diminutive b ut considerable role in high risk for metabolic diseases a s the caus e of unexpected death in infants and young children [(Lundemose et a l., 1977). Although, a link between sudd en, unexpected, infant deaths and inherited metabolic diseases was m ade almost 30 years ago, but there has been increased interest in this topic re cently (A nonymous,19 96).In particular, it has been claimed that an a bnormalit y of fatty acid β-oxidation, medium chain acyl CoA deh ydrogenase (MCAD) deficiency, could cause 3% of cases of sudd en infant death s yndrome (SIDS) (Howat et al., 1984)and inherited metabolic diseases could account, in total, for about 10% o f these deaths (Emery et al., 1 988). SIDS is the unex pected death of an apparently well infant over one month of age, for which n o cause can be found, in spite of a p ost-mortem exam ination (Beckwith,19 70). The exact c ause of such sudden unexplained death in infants u nder one year of age remains unkn own in approximately 80% of cas es (Hu nt, 2001). Of the known c auses, infections account for the highest number of ―na tural‖ causes (Platt et al., 2000; Cote et al., 1999; Sinclair-Smith et al., 1976). The diagnosis of SIDS or sudden un explained death in infancy (SU DI), still remains the largest single cause of death in children in the indu strialized world. The frequency is r eported at 1:1000 live births and rep resents 25% of all d eaths in the first year of life. Em ery was the first to monit or that a broad arra y of metabolic d isorders may present as su dden infant death syndrome(Sinclair- Smith et al., 1976). It may result from dramatic cardiac Int'l J. Clin. Biochem. 167 failure, sh ock or cardiac ar rest in many metabo lic circumstances. At least 31 m etabolic disorders are listed as causes of SIDS, ther e is some d oubt as to th e validity of some r eports (Saud ubray &Charpentier, 2000). In practice with th e exception of the fatty acid oxidation defects, the majority of these d isorders do not strictly present as SIDS but rather as an acute m etabolic crisis with clear clinical symptoms, which precedes death by hours or even a few days. The most lik ely metabolic causes o f sudden unexplained death are listed belo w: In herited d efects of fatty acid oxidation and ketogenesis. Urea cycle d isorders - most c ommonly ornithine transcarbamoylase (OTC) d eficiency. Organic acidu rias e.g. m ethylmalonic (MMA), propionic (PA) and iso valeric aciduria (IVA). C ongenital l actic acidosis i.e. pyruvate dehydrogen ase deficiency (PDH), Respirat ory chain disorders, Biotinidas e deficiency. Carboh ydrate disorders e.g. galactosaemia, glycogen storage dise ase t ype I (GSD I), hereditar y fruct ose intolerance, fructose 1,6-bisphosphatase deficiency. Jamaican Vomiting Si ckness (JVS) Ackee fruit toxicity has be en kn own since the nineteenth century and popularly called ―Jamaican vomiting sickness‖ because of the characteristic severe bouts of vomiting (Barceloux, 2009). The term ackee, is derived from ―anke‖ and ―akye-fufuo,‖ which are used to describe the acke e apple fruit commonl y found in west Africa (Grunes et al., 2 012; Atol ani et al., 20 09). It is the national fruit in Jamaica w here the toxicity is endemic (Barceloux, 2009; Emanuel & Benkeblia, 2012) 56,59]. Ackee fruit is known scientifically as B lighia sapida belonging to the sapindaceae family (Atolani O et al., 2009). Ackee frui t poisoni ng i s caused by i ngestion of the unripe arils of the ack ee fruit, its seeds, and husks induc es severe hypoglycemia presumably as a result of inhib iting fatty acid oxidation. The stu dy of hypoglycin, which causes J amaican vomiting sickness in humans (Meda et al., 1999; Oludolapo et al.,2015) stim ulated an int erest in in hibitors of fatty acid oxidation. In animals, hyp oglycin is metabolized by deamination and oxidative decarboxylation to meth ylenecyclopropylacet yl-CoA, which inactivates s everal acyl-CoA deh ydrogenases and thereby inhibits β-oxidati on (J oskow et al., 2006). Hassel and Reyle in 195 4 first isolated th e two to xic constituents, hypoglycin A and B from the arils and se eds of the unr ipe ackee (Hassal &Reyle, 1955) resp ectively which in hibits the acyl CoA dehydrog enase and thus, b eta oxidation of fatty a cids is blocked, lea ding to various complications (Satyanarayana, 3 rd edn) b ut mainly induce severe hypoglycemia. Hyp oglycin A is met abolized by the liver to methylene c yclopropyl acetic acid, a toxic m etabolite th at inhibits the transp ort of long-ch ain fatty acids int o mitochondria, suppressing their oxidation. This impairs gluconeogenesis r esulting in hypoglycemia after glycogen stores are exh austed. Hypoglycin A also inh ibits the dehydrogenation o f s everal acyl-coenzyme A, causing an accumulation of serum fatty ac ids (Tanaka et al., 1 976). Hepatotoxicity that may oc cur is r elated to the metabolites of the toxin while CNS manifestations are attributable t o d irect to xic effect and h ypoglycemia. The unripe ackee fruit contain s hypoglycin A in a concentration 100 times higher than t hose in the ripe ackee fruit, whereas h ypoglycin B found only in the seeds of the fruit has a less-potent h ypoglycemic activity than A (Golden & Williams, 20 02; Kean & Hare, 1980). It is ch aracterized by acut e gastrointestinal i llness and hypoglycemia. In severe cases, central nervous system (CNS) depressi on can also occur. Toxicit y is dos e dependent and usually manifests wi thin 6–48 hours of ingestion wi th r ecovery usually within 1 w eek (Meda et al., 1 999). Symp toms begin with and then subsequentl y more vomiting, seizures, and coma. I n fatal c ases, d eath usually occurs within 48 h ours of ingestion(Meda et a l., 1999; Oludolapo et al., 20 15; Joskow et al., 2006). Clinical mani festations of Jamaica n vomiting sickness are: (Meda et al., 1 999; Oludolapo et al., 2015; Joskow et al., 20 06). Hypogl ycemia Hepatic in jury A ciduria Clinical fea tures of FOAD Deficiency There may be a range of clinical pre sentations ranging from mild liver dysfunction, cardiomyopathy and/or skeletal myopathy to se vere liver disease that may present with a recurrent Reye-like syndrome that may start in th e infantil e period with h epatic steat osis, unexplained hep atic failure and n on-ketotic hypogl ycemia (Rudolph &Rudolph, 2011). Stresso rs such as fa sting may exacerbate the hepatic dis ease. General ma nifestations E xtreme sleepiness B ehavior changes Irritable m ood E nlarged heart Heart failure Fe ver Naus ea/Vomiting Diarrhoe a Decre ased Appetite Hypogl ycemia Muscle weakness Specific Manifestations Long Chain Fatty Acid Transport/Binding Effect E pisodic acute liver failure Prakash 168 En cephalopathy Hyperam monemia Carnitine Palmitoyl transferase I (CPT-I ) Defic iency (Gempel et al., 2002) Fasting Intolerance Hypok etotic Hypoglycemia S eizures Coma Carnitine P almitoyl transferase II(CPT-II) Deficiency (Gempel et al., 2002) Classic m uscle form:Fasting/Stress—E pisodic Myoglobinuria Se vere ne onatal form: Hypoketotic H ypoglycemia, heart, liver,seizures/coma Carnitin e –Acyl Carnitine Translocase (CACT) Deficiency(Angelini , 1992) Neonat al form: Cardiac Fasting intol erance—Coma Mild Form: Hypoglycemia Very Long Chain A cyl-CoA Dehy drogenase (VLCAD) deficiency C typ e: Cardiac, sudden death H typ e: Hepatic, episodic hypoketotic h ypoglycemia Long Chain Acyl-CoA Dehydroge nase (LC HAD) Deficiency Fasting—E pisodic hypoketotic hyp oglycemia Maternal HEL LP Syndrome Muscle, he art, Liver and Eyes Medium Cha in Acyl-CoA Dehydrogenase (M CAD) Deficiency Fasting--- Episodic hypoketotic hyp oglycemia Vom iting Acid osis Coma Short Chain Acyl-CoA Dehy drogenase (SCAD) Deficiency Neonat al: Vomiting Acidosis, Devel opmental delays Chronic: Musc le Weakness Diagnosis Prenatal Diagnosis This diagnosis is made by biochemical or molecular methods following chorionic vil lus s ampling or amniocentesis. The m ost preferred technique is m ut ation analysis, if the molecular defect i s known i n the index case (Vance & Vance, 2002). With history of matern al liver d isease with complicating pregnancies, prenatal diagnosis becomes comp ulsory (Shekhawat et al., 20 05). Diagnosis in Newborns Acylcarnitine profiling of b lood sp ots using tand em mass spectrometry is th e screening techniqu e for new born cases. T he de termination of blood a cylcarnitine profiles by tandem mass sp ectrometry from filter pap er bl ood spots allows detection of f atty acid oxidation disorders caused b y deficienci es of MCAD, VLCAD, LCHAD/TFP, ETF/ E TF-DH, SCHAD, SCAD and HMG-CoA lyase. In CPT-1 deficiency, total carnitine l evels are incr eased (150–200% of n ormal) (Stanley et al., 1992). In all of t he other de fects, e xcept HMG-CoA synthase d eficiency, total carnitine levels are reduced to 25-50% of normal (secondary carnitine d eficiency). Thus, simp le measurement of plasm a total carnitine is o ften helpful to determine the presence of a fatty acid oxidation dis order. It should be emphasiz ed that samples mu st be t aken in the well-fed state with normal dietary carnitine intake because patients with disorders of fatty acid o xidation may sh ow acute increases in the plasma total carnitine during prolonged fasting or du ring attacks of illness. The following FAO Ds are diagnosed by ne wborn screening: CAC T deficiency CPT II de ficiency (neonatal and lat e onset) V LCAD deficiency MCAD de ficiency S CAD d eficiency and a few other d isorders like electron transport flavoprotein-ubiquinon e oxidoreductase (ETF-QO) d eficiency, α-ETF deficiency and β-ETF deficiency (Sim et al., 2002) Diagnosis in Children and Adults Fatty acid transport st udies u sing fibrobl asts reveal possible f atty acid transporter defects. Li ver biopsy may be necess ary if patient present with p rimarily hepatic dysfunction and m ay reveal st eatosis (Rudolph & Rudolph, 2011). The diagnosis of FA ODs even postmortem may help in g enetic counseling and evaluation of siblings (Blau N, 2014). Blood and urine samples collected i mmediately prior to treatment of an acute episode of illness can b e used for this purpos e, e.g. by showing elevated plasma free fatty acid bu t inappropriately low ketone levels at the time of hypoglycemia. The main laboratory studies include r outine labs such as:  Comp lete blood count (C BC) B asic metabolic panel (BMP) Hepatic p anel A mmonia L actate Creatin e phosphokinase (CPK) A cylcarnitine levels MS/MS analysis of organic acids, Int'l J. Clin. Biochem. 169 Plasma c arnitine and u rine acylglycine analysis, wi th a definitive diagnosis based on m utation analysis or m easurement of specific enzyme activity (Sim et al.,2002). Table 2: Fatty acid-oxid ation disorder s with distingu ishing metab olic markers Disorder Plasma acylcarnit ines Urinary acylgl ycines Urinary organic a cids VLCAD Tetradecenoyl- MCAD Octanoyl- Hexanoyl- Decenoyl- Suberyl- Phenylpropionyl- SCAD Butyryl- Butyryl- Ethylmalon ic LCHAD 3-Hydroxy-palm itoyl- 3-Hydroxydicarbo xylic 3-Hydroxy-oleo yl- 3-Hydroxy-linol eoyl- DER Dodecadienoyl- ETF and ETF-DH Butyryl- Isovaleryl- Ethylmalon ic Sovaleryl- Hexanoyl- Glutaric Glutaryl- Isovaleric HMG-CoA lyase Methylglutaryl- 3-Hydroxy-3-m ethylglutaric DER, 2,4-dieno yl-coenz yme A reduc tase; E TF, electron-transfer fla voprotein; E TF-DH, E TF dehydrogenas e; HM G- CoA, 3-hydro xy-3-methylglutar yl-coenzym e A; MCAD, m edium-chain acyl-coenz yme A dehydrogenas e; SCAD, short- chain acyl-co enzyme A de hyd rogenase;VL CAD, very-long-chain acyl-c oenzyme A deh ydrogenase In vitro Cultured skin fibroblasts or lymphoblasts from patients can also b e used to demonstr ate a g eneral defect in fatt y acid oxidation using 14C or 3H-labeled su bstrates. In addition, different chai n-length fatty acid substrate s can be used wi th these cel ls to localize the pr obable site of defect. Tandem mass sp ectrometry using deuterated stable isotopes fatt y acids has become an im portant method for in vitro t esting in cultured cells. In the hepatic presentation of an y of the fatty acid o xidation disorders, a liver bi opsy obtained during a n acut e episode of illness shows an increase in neutr al fat deposits which m ay have either a micro- or m acrovesicular appearance (Sim et al., 2002; Vishwanath, 2016). Enzyme Assays Cultured skin f ibroblasts or cultured lymphoblasts have become the p referred m aterial in which to m easure the in vitro activities of spec ific steps in the fatty acid oxidation pathway. All of the known defects, except HMG-C oA synthase, are express ed in these cells and results of assays in cells from both control a nd affected patients have be en rep orted. Because th ese assays are not widely available, they are most usefully applied to con firm a site of defect that is suggested by o ther clinical a nd laboratory data (Vish wanath, 2016) Treatment a nd Management Acute Illnesses When p atients with fatty ac id oxidati on disorders become ill, treatment w ith intravenous glucose should be given immediately. Delay may result in sudden death or permanent brain damage. The goal is to pro vide sufficient glucose to stimulat e insulin secretion to levels th at will not only suppress fatty acid oxidation in li ver and mu scle, but also b lock adipos e ti ssue lipol ysis. S olutions of 10% dextrose, rather t han the usual 5%, should be used at infusion rates o f 1 0 mg/kg per min or greater to main tain high to normal lev els of plasma g lucose, above 100 mg/dl (5.5 mmol/l). Res olution of coma m ay not be imm ediate, perhaps bec ause of the toxic effects of fatty acids for a few hours in m ildly ill patients or as long as 1–2 days in severely ill patients (The Philadelphia Guide, 2005) Long-Term Therapy The go al would be to st op f at cat abolism by pre venting further f atty ac id oxidation. The initial steps would be the prevention of h ypoglycemia in p eriods of catabolic stress by using frequent feeds a nd clinical supervision durin g periods o f illnesses. A low fa t, high carbohydrate diet is recommended. Dietary f at restriction is not indicated in MCAD deficiency and mild long-chain FAODs recently identified by newborn screening. L ong chain fat, h owever, Prakash 170 needs to be restricted in severe long chai n FAODs and substituted by medium-chain triglycerides . Hospital admission is recommended for procedures that would require the patients to t ake nothing orally for >8 h, especially if less than 1 year of age. Carnitine is undisputedly e ffective in patients with carnitine transporter de ficiency (Bl au N et al., 2014). Li ver transplantation may b e th e ultim ate consid eration if there is no evidence o f neurological disease or other systemic involvement that m ay im pair recovery and return to baseline funct ion (Vishwanath, 2016; An gelini et al., 2011). Other Therapy Since medium-chain fatty acids bypass the c arnitine cycle and enter the midportion of th e m itochondrials- oxidation spir al directl y, it is possible that they might be used as fuels in d efects which block either th e carnitine cycle or long-chain s-oxidation. For examp le, dietar y MCT was su ggested to be helpful in a p atient with LCHAD d eficiency. The b enefits of MC T ha ve not been thoroughly in vestigated, but MCT cle arly m ust not be used in patients with MCA D, SCAD, SCHAD, ETF/ETF- DH, HMG-CoA synthas e, or HMG-CoA lyas e deficiencies. Some patients with mild variants of ETF/ETF-DH and SCAD deficiencies have been r eported to r espond to supplementation with high d oses o f riboflavin (100 mg/d ay), the cofactor for thes e enzymes. Triheptanoin was su ggested to be o f benefit I th ree cases of VLCAD as an anaplerotic substrate, but h as not ye t been confirmed by contro lled studies (Roe et al., 2002). Prognosis Although acute episodes carry a h igh risk of m ortality or permanent brain damag e, many patients with disorders of fatty acid oxid ation can b e easil y managed b y avoidanc e of prolonged fasts. These patients h ave an excellent long-term prognosis. Patients with ch ronic cardiomyopathy or sk eletal mu scle weakness have a more guarded p rognosis, since the y seem to have more severe defects in fatty acid o xidation. F or example, TRANS or t he severe variants of CPT-2 and ETF/ETF- DH deficienci es frequentl y lead to death in t he newborn period. On the other h and, the mild for m of CPT-2 deficiency may remain silent as long as pa tients avoid exercise stress (Sim et al., 2002; V ishwanath, 2016) CONCLUSION The oxidation of fatty acids in mitoch ondria plays an important role in energy metabolism and genetic disorders o f this p athway ma y cause metabolic diseases. Enzyme deficiencies can block the m etabolism at d efined reactions in t he mitochondri on and l ead to accu mulation of speci fic substrates causing severe c linical manifestations. This review completes lucidly with the fundamentals of the pathway of mitochondrial β- oxidation, control of p athway flux, FA ODS and its clinic al manifestations. ACKNOWLEDGEME NT Author would lik e to thank Ms. Khushb u Yadav, Medical Microbiologist and Lecturer, Nepal Dental Hospital and Research I nstitute, Janakpurdham, Nepal for her valuable posit ive critics during preparation of this manuscript. 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Inpati ent Pediatrics. Philadelph ia, Lippin cott Williams &Wilkins. Thorpe C and Kim JJ (1995). Structure and mechanism of action of the acyl-CoA d ehydrog enases. FASEB J. 9:718 725. U. Satyanarayana,U. Chakrapani, Biochemistry by U.Satyanarayana. 3 rd Edit ion. Prakash 172 Vance DE and Vance JE (20 02). Biochem istry of Lip id Lip oproteins and Membran e Oxidation of fatty acids in e ukaryot es (4th Edn.) Elsevier Science B.V. Vishw anath AV (2016). Fat ty Acid Beta-Oxi dation Disorders: A B rief Review An n Neurosci. 23:51–55 Wanders RJ, Vrek en P, den Boer ME, Wijb urg FA, van Gennip AH, IJlst L (1999). D isorders of mitochondrial fatty acyl-CoA betaoxidation. J Inherit Metab Dis. 22:442–487. Wanders RJA, Vrek en R, Ferdinand usse S, Jansen GA, Wate rham HR, Van Roermu n CWT and Van Grunsven EG (2001). Peroxisom al fatty acid and beta- oxidation in humans: enzym ology, peroxisomal metabol ite transp orters and peroxis omal diseases. Biochem Soc Trans. 29: 25 0-267. Yokota I, Indo Y, Coates PM, Tan aka K (1990). M olecul ar basis o f medium c hain acyl-coenzyme A dehydrogenase deficien cy. An A to G t ransition at po sition 985 that causes a lysin e-304 t o g lutam ate substitu tion i n the m ature protein is the singl e prevalen t mutation. J Clin Invest. 8 6:1000–1 003. Citations (10) References (67) ... The stearate biosynthesis pathway produces stearate, a type of saturated fatty acid, that is incorporated into various lipids (Wakil 1961). In contrast, the fatty acid oxidation pathway is the metabolic process through which fatty acids are broken down to produce energy (Prakash 2018). This dueling upregulation of pathways involved in the synthesis and metabolism of fatty acids suggests a destabilization of cellular metabolism, a phenomenon also observed in MASH. ... Identification of Four Mechanisms for Per- and Polyfluoroalkyl Substances (PFAS) Through Transcriptomic Profiling from 24 PFAS-Exposed Human Liver Spheroids Article Full-text available Jun 2025 TOXICOL SCI Gregory Addicks Andrea Rowan-Carroll Karen Leingartner Ella Atlas Per- and polyfluoroalkyl substances (PFAS) are persistent and widespread contaminants. Epidemiological effects of PFAS include increased serum cholesterol, decreased immune response to vaccination and disease, and increased incidence of cancer; however, PFAS modes of action remain unclear. Herein, we analyzed gene expression data from human liver spheroids that were exposed to several concentrations of 24 different PFAS. Benchmark concentration (BMC) response modeling was used to identify the 250 lowest gene BMCs for each PFAS. Hierarchical clustering analysis revealed four functionally diverse gene sets. Each gene set was affected by a distinct group of PFAS, while individual PFAS were usually part of more than one PFAS group. The biological roles of these gene sets relate to: 1) cholesterol biogenesis and cholesterol clearance (downregulated by 7 fluorocarbon or longer PFAS), putatively through discordance of cholesterol sensing by SCAP and LXR due to membrane integration of PFAS; 2) lipolysis (upregulated by 8 carbon or shorter PFAS); 3) innate immunity (downregulated by most PFAS); and 4) adaptive immunity (downregulated by sulfonate type PFAS). The distinctions between the four PFAS groups suggests that PFAS can act through at least four independent mechanisms. The molecular characteristics of each PFAS group may by useful for understanding the molecular interactions leading to their effect on gene expression. That some PFAS congeners are included in more than one PFAS group suggests that individual PFAS can act through multiple unrelated molecular interactions. This transcriptomic analysis offers a major advancement to the understanding of the molecular mechanisms underlying the effects of PFAS exposure, and provides guidance for future work that may strengthen links between PFAS exposure and their proposed effects on human health. View Show abstract ... Further elongation and desaturation processes contribute to the synthesis of cholesterol, a critical substrate for steroid hormone production. These pathways highlight the multifaceted roles of palmitic acid in both cellular energy homeostasis and the biosynthesis of biologically active lipids 36,37 . Hexadecenoic acid is known to influence reproductive physiology by modulating prostaglandin synthesis, improving oocyte quality, and enhancing the uterine environment. ... Metabolomic profile of dromedary camel follicular fluid during the breeding and non-breeding seasons Article Full-text available Mar 2025 Ahmed Abdoon Seham Samir Soliman Noha S. Hussein Abdel-Hamid Z. Abdel-Hamid Understanding the metabolic profile within the follicular microenvironment is crucial for optimizing reproductive efficiency in camels. In this study, we examined the metabolomic profile of camel follicular fluid (FF) during the breeding (n = 10) and non-breeding seasons (n = 10). Gas chromatography-mass spectrometry (GC-MS) was utilized to describe the metabolites present in follicular fluid samples. The results found considerable differences in the metabolomics profiles between the breeding and non-breeding seasons. Hexadecenoic acid, galactose and glucose levels were significantly (P < 0.05) higher in camel FF during the breeding season, while 9-octadecenamide, oleonitrile, glycine, octadecanamide, cholesterol, and propanoic acid were higher (P < 0.05) in FF during the non-breeding season. Multivariante analyses pointed to those 9 metabolites, and univariate analysis showed hexadecenoic acid, galactose, glucose, and oleanitril were the most significant ones in camel follicular fluid collected during both breeding and non-breeding seasons. The univariate and multivariate analyses showed an increase in the levels of hexadecanoic acid, galactose, glucose, and a depletion in the level of oleanitrile in the breeding season compared to the non-breeding season. The ROC curve and statistical analysis showed that hexadecanoic acid, galactose, and oleanitril with AUC = 1 were promising to be seasonal biomarkers of fertility in female camels. In conclusion, the metabolomic analysis of camel FF reveals distinct changes in metabolite levels between breeding and non-breeding seasons, reflecting adaptive metabolic responses to support reproductive processes. These results offer valuable insights into the reproductive physiology of camels and offer practical implications for potential biomarkers and assessing the reproductive status in camels, which can be utilized in reproductive management and conservation efforts in these valuable animal species. View Show abstract ... The stearate biosynthesis pathway produces stearate, a type of saturated fatty acid, that is incorporated into various lipids (Wakil 1961). In contrast, the fatty acid oxidation pathway is the metabolic process through which fatty acids are broken down to produce energy (Prakash 2018). This dueling upregulation of pathways involved in the synthesis and metabolism of fatty acids suggests a destabilization of cellular metabolism, a phenomenon also observed in MASH. ... Deciphering PFAS Mode of Action: Comparative Gene Expression Analysis in Human Liver Spheroids Article Full-text available Mar 2025 TOXICOL SCI Andrea Rowan-Carroll Matthew J Meier Carole Yauk Ella Atlas Understanding the mechanisms by which environmental chemicals cause toxicity is necessary for effective human health risk assessment. High-Throughput Transcriptomics (HTTr) can be used to inform risk assessment on toxicological mechanisms, hazards, and potencies. We applied HTTr to elucidate the molecular mechanisms by which Per- and Polyfluoroalkyl Substances (PFAS) cause liver perturbations. We contrasted transcriptomic profiles of PFOA, PFBS, PFOS, and PFDS against transcriptomic profiles from established liver-toxic and non-toxic reference compounds, alongside peroxisome proliferator-activated receptors (PPARs) agonists. Our analysis was conducted on metabolically competent 3-D human liver spheroids produced from primary cells from 10 donors. Pathway analysis showed that PFOS and PFDS perturb many of the same pathways as the known liver-toxic compounds in the spheroids, and that the cholesterol biosynthesis pathways are significantly affected by exposure to these compounds. PFOA alters lipid metabolism-related pathways but its expression profile does not closely match reference compounds. PFBS upregulates many degradation-related pathways and targets many of the same pathways as the PPAR agonists and acetaminophen. Our transcriptional analysis does not support that these PFAS are DNA damaging in this model. A multidimensional scaling analysis revealed that PFOS, PFOA, and PFDS cluster together in the same multidimensional space as liver-damaging compounds; whereas, PFBS clusters more closely with the non-liver-damaging compounds. Benchmark concentration-response modeling predicts that all the PFAS are bioactive in the liver. Overall, our results show that these PFAS produce unique transcriptional changes but also alter pathways associated with established liver-toxic chemicals in this liver spheroid model. View Show abstract ... Image made with biorender.com and adapted from [23,24]. ... The associations between prenatal plastic phthalate exposure and lipid acylcarnitine levels in humans and mice Article Full-text available Jan 2025 REPROD TOXICOL Kristina Vacy Thusi Rupasinghe Alicia Bjorksten Anne-Louise Ponsonby Phthalates are ubiquitous environmental pollutants known for their endocrine-disrupting properties, particularly during critical periods such as pregnancy and early childhood. Phthalates alter lipid metabolism, but the role of prenatal exposure on the offspring lipidome is less understood. In particular, we focused on long chain acyl carnitines - intermediates of fatty acid oxidation that serve as potential biomarkers of mitochondrial function and energy metabolism. This study aimed (i) to investigate the association between prenatal phthalate exposure and the child’s blood acylcarnitine concentrations and, (ii) to evaluate the impact of prenatal administration of di-(2- ethylhexyl) phthalate (DEHP) on acylcarnitine levels in mouse offspring blood, brain and liver. We conducted analyses of both a prospective birth cohort study and an experimental study in mice. From the Barwon Infant Study cohort (1074 mother-child pairs), prenatal phthalate exposure was assessed at 36 weeks’ gestation and its association with acylcarnitine levels was examined in cord blood, and child’s blood at 6 months, 12 months and 4 years. In mice, pregnant C57BL/6 J mouse dams were exposed to 20 μg/kg DEHP for 5 days mid-gestation, and offspring tissues were analyzed at 1 month of age postnatally for acylcarnitine profiles. Our findings demonstrate that prenatal phthalate levels (specifically butyl benzyl phthalate (BBzP) and diisobutyl phthalate (DiBP)) are inversely associated with total long chain acylcarnitine levels in human cord blood at birth. In contrast, BBzP was positively associated with the long chain acylcarnitines at 12 months of age. In mice, prenatal DEHP exposure for only 5 days led to decreased palmitoylcarnitine (AC16:0) levels in the brain and liver, but not in blood. Taken together, our findings highlight that prenatal phthalate exposure can alter acylcarnitine profiles, indicating disruptions in fatty acid metabolism that may have long-term effects on metabolic health. View Show abstract ... Because ketoacids are strong acids and this occurs systemic acidosis. Since ketoacids are excreted from kidney, urinary electrolyte loss is facilitated and osmotic diuresis caused by hyperglycemia worsens (Kerner & Hoppel, 2000;Prakash, 2018). ... DIABETES MELLITUS in CATS: An Update On Physiology, Etiopathogenesis, Clinical Diagnosis, Treatment Chapter Full-text available Dec 2022 Erman Koral Mutlu Sevinc Diabetes mellitus is defined as persistent hyperglycemia caused by relative or absolute insulin deficiency. Insulin deficiency occurs when beta (β) cells are destroyed or their functions are impaired. Diabetes mellitus (DM) is one of the two most common endocrinopathy among feline diseases. Produced insulin from beta cells of the pancreas, promotes cellular glucose uptake for the energy needs of most cells in the body. Diabetes mellitus in cats is clinically and pathologically quite similar to type 2 diabetes mellitus in humans. Weight gain significantly reduces insulin sensitivity, and each kilogram increase in body weight reduces insulin sensitivity by 30%. Symptoms of diabetes mellitus are hyperglycemia, glycosuria, polyuria, polydipsia, polyphagia, weight loss, anorexia, cataracts in dogs, neuropathy in cats. Diabetes mellitus is usually diagnosed based on the determination of persistent fasting hyperglycemia and glycosuria with clinical findings. However, these alone are not enough and glycosylated proteins such as fructosamine and glycosylated hemoglobin are used markers for advanced diagnostic test. In recent years, the goals in the treatment of clinical diabetes have changed, The main aim of treatment is to regulate blood glucose concentration, to eliminate clinical manifestations of the disease and to avoid complications that may occur during treatment, especially hypoglycemia. However, in recent years, new approaches such as GLP-1, GIP, cabergoline and lispro insulin will shed light on future research for the treatment of cats with diabetes. View Show abstract Role of fatty acids in milk production Chapter Jan 2025 Dapinder Singh Amarjeet Bhavna Jha Abhishek Pathak View Pengembangan Akuakultur Berbasis Bioflok untuk Peningkatan Produktivitas dan Keberlanjutan Book Full-text available Oct 2024 Deswati Deswati Joko Sutopo Wiya Elsa Fitri Adewirli Putra Buku Pengembangan Akuakultur Berbasis Bioflok untuk Peningkatan Produktivitas dan Keberlanjutan Industri akuakultur terus berkembang dengan pesat seiring dengan meningkatnya permintaan global akan produk perikanan. Akuakultur atau budi daya organisme air seperti ikan, udang, lobster, dan berbagai jenis organisme air lainnya, telah menjadi komponen kunci dalam memenuhi kebutuhan akan pasokan pangan protein hewani yang semakin meningkat seiring pertumbuhan populasi dunia. Akuakultur tidak hanya memiliki potensi untuk memberikan pasokan pangan yang berkelanjutan, tetapi juga dapat mengurangi tekanan pada sumber daya perikanan alami yang semakin terancam. Namun, seiring dengan pertumbuhan industri akuakultur, muncul pula sejumlah tantangan dan masalah yang perlu diatasi. Salah satu masalah yang seringkali menjadi hambatan dalam budi daya ikan dan organisme air adalah manajemen kualitas air dan pengelolaan limbah organik yang dihasilkan oleh organisme yang dibudidayakan. Kualitas air yang buruk dapat mengancam kesehatan ikan, menghambat pertumbuhan, dan bahkan menyebabkan kerugian finansial yang signifikan bagi petani akuakultur. Dalam menghadapi tantangan tersebut, teknologi bioflok telah terbukti sebagai salah satu solusi yang inovatif dan efektif dalam mengelola limbah serta meningkatkan efisiensi budi daya. View Show abstract AQUATIC PLASTICS WASTE BIODEGRADATION USING PLASTIC DEGRADING MICROBES Article Full-text available Dec 2021 Angga Puja Asiandu Agus Wahyudi Septi Widiya Sari Plastic is a synthetic polymer that is highly used every year in almost every field of life. Aquatic plastic waste is plastic that scattered in the aquatic environment in the form of macroplastic or microplastic. Plastic production is expected to continue increasing from year to year. Due to the massive production and use of various plastic products, the accumulation of plastic waste in the environment is still increasing resulting in environmental pollution. Biodegradation is considered as the appropriate method to solve the problem. Plastics biodegradation involves various enzymes produced by plastic degrading microorganisms including algae, bacteria, and fungi. During biodegradation, plastics polymer will be converted into microbial biomass and gases through several steps including biodeterioration, biofragmentation, depolymerization, assimilation and mineralization. Thus, this process has less side effect on the environment. View Show abstract Fatty Chain Acids Risk Factors in Sudden Infant Death Syndrome: A Genetic Algorithm Approach Chapter Oct 2020 Lect Notes Comput Sci Karen Villagrana Laura A. Zanella Calzada Irma Gonzalez-Curiel Carlos Eric Galván Tejada Medicine and artificial intelligence (AI) have made great progress, since they have achieved unprecedented knowledge and explanations about how the human body works and about some diseases that seemed to have no way of preventing, diagnosing or treating or simply help make those processes more efficient. Sudden infant death syndrome (SIDS) could benefit from AI, since to date it has not been possible to clarify what actually causes it, devices to monitor vital signs have been used so far, an recently was made a predictive model to predict results from an autopsy in infants, however, further investigation is necessary to more effectively prevent. The main objective of this work is to be able to find a set of factors related to short chain fatty acids (SCFA) that could help to understand the risk of SIDS. Was used a public dataset named “Analysis of SCFA profile in infants dying of SIDS compared to infants dying of controls”, that contained SCFA values, of deceased children, labels them as SIDS death and from another cause (control). For pre-processing some variables were removed from the dataset. An analysis was performed with a feature ranking with genetic algorithm (GA) and risk analysis using the information of SCFAs and their relationship with SIDS, is presented. The median was calculated for each of the SFCA, which served to form two groups necessary to evaluate the risk difference and the risk relationship depending on the amount of acids present for each subject. As results, Octanoic acid represents a risk difference of 18% for the population with an amount less than 3.5 uM and an individual risk of 1.28 times. On the other hand, hexanoic and propinoic acids present a risk difference of less than 11% with a lesser amount of 23 and 128 uM respectively, as well as an individual risk of approximately 0.85 times. As conclusion there is 18% higher risk of developing SIDS if octanoic acid is less than 3.5 uM or 1.28 times greater risk. On the other hand, with hexanoic and propionic acid, they agree that there is an 11% lower risk of developing SIDS (of manner independent), if the values are less than 23 uM and 128 uM, respectively. View Show abstract Real-time diagnosis of sentinel lymph nodes involved to breast cancer based on pH sensing through lipid synthesis of those cells Article Full-text available May 2020 Zohreh Sadat Miripour Parisa Aghaee Fereshteh Abbasvandi Mohammad Abdolahad Lipid synthesis is the recently found metabolism of cancer cells after their metastasis to Lymph Nodes (LNs). Carbonic acid is the main byproduct of the lipid metabolism in such cells which resulted in acidification of LN ambient. Hence calibrated pH sensing could be a diagnostic method to find involved LNs. Here we designed a simple pH sensing method by a syringe containing sterile PBS and embedded by litmus paper to intraoperatively check the pH of LN fluid. Injected Phosphate Buffer Saline (PBS) would homogenize the LN fluid and litmus needle would detect the pH of the LN. We presented an experimental pathological calibration for the pH values in correlation with cancerous states of the LNs. This system named Metabolism based Metastatic Lymph Diagnoser (MMLD) could be a real-time noninvasive tool for precise and fast diagnosis of involved LNs. View Show abstract Show more Ackee Fruit Poisoning in Eight Siblings: Implications for Public Health Awareness Article Full-text available Aug 2015 Oludolapo Sherifat Katibi Rasaq Olaosebikan Mohammed Baba Abdulkadir Rukayat Murtala Ackee apple fruit is a native fruit to Jamaica and some parts of west Africa. Its toxicity known as "Jamaican vomiting sickness" dates back to the nineteenth century. However, there is a dearth of reported published data on toxicity from Nigeria where it is popularly known in the southwest as "ishin." We report a case series of eight previously well Nigerian siblings who presented at various intervals after ingestion of roasted seeds and aril of the ackee fruit. © The American Society of Tropical Medicine and Hygiene. View Show abstract Adding Value to Tropical Fruits – The Case of the Jamaican Ackee Industry: Lessons for Policy and Practice. Chapter Full-text available Aug 2012 Machel A. Emanuel Noureddine Benkeblia Tropical fruits offer a significant opportunity for agricultural and economic growth for many ACP countries. However, while production, processing and marketing of some better known fruits such as citrus, mangoes, avocadoes and bananas, has benefitted from significant investments including in research and development, primarily to service export markets; this has not been the same for many other tropical fruits. This dossier comprises two lead articles by ACP and EU experts and provides links to relevant documentary resources on tropical fruit processing. It seeks to highlight the challenges and opportunities in adding value to tropical fruits and provides policy guidelines to support industry development. In the first lead article, Ludovic Temple of CIRAD, France, explains how the development of the horticultural sector and the determinants of innovation are linked. For major tropical fruits that are traded internationally; foreign investment, research and development and technological advances e.g. refrigerated storage as well as improvements in logistics have been critical. International standards have also driven innovation, and emerging standards such as organic and fair trade are creating new market opportunities. While the benefits have accrued to large-scale well-organized producers and other actors along the value chain, small family-scale fruit production and processing enterprises have not been able to meet the requirements of large-scale retailers. They are also reluctant to take the risks of investing in new technology, such as introducing novel disease-resistant varieties. There is still a need to accelerate the adoption of new technologies and improve coordination to their benefit. In the second lead article, Machel A. Emanuel and Noureddine Benkeblia, University of the West Indies, Jamaica, demonstrate the challenges faced in developing the Jamaican ackee (Blighia sapida K.D. Koenig) industry. Ackee is well accepted by Jamaicans but not well known in many other countries; it has the potential to contribute to the growth of the Jamaican economy if the industry can increase its penetration of international export markets. However, the ackee contains a toxic compound, hypoglycin A, which poses a challenge in meeting food safety requirements. There are also other challenges hindering industry development including availability of suitable processing variety in adequate quantities. Research and technology development have supported industry development but challenges remain. In 1972, the US Food and Drug Administration effectively banned the imports of ackee from Jamaica. An upper limit of 100 mg/kg of hypoglycin A was set. In 1990 an accurate detection procedure was developed, allowing importation of tested produce into the USA to begin again. However, in 2005, the exportation of ackee to the US market was again suspended for almost one year for technical reasons. Presently, only certified agro-processors with food safety controls in place can export canned ackee that will not be automatically detained. Further growth of the industry needs a concerted effort of scientists, engineers, policymakers, investors and entrepreneurs including farmers to support this export-oriented industry which remains heavily dependent on maintaining the confidence of the overseas consumers. Tropical fruit industry expansion in ACP countries does not rely only on coordinating and improving production and logistics efficiencies but requires a systems approach. This folder was compiled and edited by CABI and CTA, June, 2012. View Show abstract Fulminant Hepatic Failure Attributed to Ackee Fruit Ingestion in a Patient with Sickle Cell Trait Article Full-text available Oct 2012 Dianne E Grunes Irini Scordi-Bello Matthew Suh Swan N. Thung We report a case of fulminant liver failure resulting in emergent liver transplantation following 3 weeks of nausea, vomiting, and malaise from Jamaican Vomiting Sickness. Jamaican Vomiting Sickness is caused by ingestion of the unripe arils of the Ackee fruit, its seeds and husks. It is characterized by acute gastrointestinal illness and hypoglycemia. In severe cases, central nervous system depression can occur. In previous studies, histologic sections taken from patients with Jamaican Vomiting Sickness have shown hepatotoxicity similar to that seen in Reye syndrome and/or acetaminophen toxicity. We highlight macroscopic and microscopic changes in the liver secondary to hepatoxicity of Ackee fruit versus those caused by a previously unknown sickle cell trait. We discuss the clinical variables and the synergistic hepatotoxic effect of Ackee fruit and ischemic injury from sickled red blood cells, causing massive hepatic necrosis in this patient. View Show abstract Observations on the pathological anatomy of the sudden infant death syndrome Article Jan 1970 J.B. Beckwith View Fatty Acid Beta-Oxidation Disorders: A Brief Review Article Mar 2016 Vijay A. Vishwanath Background Mitochondrial fatty acid β-oxidation disorders (FAODs) are a heterogeneous group of defects in fatty acid transport and mitochondrial β-oxidation. They are inherited as autosomal recessive disorders and have a wide range of clinical presentations. The background information and case report provide important insight into mitochondrial FAODs. The article provides a wealth of information describing the scope of these disorders. Key Messages This article presents a typical case of medium chain acyl-CoA dehydrogenase deficiency and summarizes the pathophysiology, clinical presentation, diagnosis and treatment of mitochondrial FAODs. View Show abstract Der Abbau aromatischer Fettsäuren im Tierkörper Article Jan 1905 F. Knoop View Physician's Guide to the Diagnosis, Treatment, and Follow-up of Inherited Metabolic Diseases Article Jan 2014 Nenad Blau M Duran K M Gibson Carlo Dionisi-Vici View Beta-oxidation of unsaturated fatty acids: a revised pathway Article Dec 1987 TRENDS BIOCHEM SCI Horst Schulz W H Kunau A revised pathway (the reductase-dependent pathway) by which polyunsaturated fatty acids are β-oxidized is presented. This pathway requires the involvement of a NADPH-dependent 2,4-dienoyl-CoA reductase in addition to Δ3-cis-Δ2-trans-enoyl-CoA isomerase and the enzymes necessary for the oxidation of saturated fatty acids. The original pathway (the epimerase-dependent pathway) with 3-hydroxyacyl-CoA epimerase as an auxillary enzyme is inoperative in mitochondria but may play a minor role in non-mitochondrial β-oxidation systems. View Show abstract Lipid Biochemistry An Introduction Book Jan 2002 Michael I. Gurr John L. Harwood Keith Frayn View γ-Glutamyl transpeptidase of the ackee plant Article Dec 1980 Eccleston A. Kean Ester Rose Hare The enzyme γ-glutamyl transpeptidase was purified from seeds of immature ackee fruit (Blighia sapida; Sapindaceae) by salt fractionation and gel filtration on Biogel P-10 and P-200. The procedure, which differs from an earlier one applied to kidney bean fruit, achieves 9.8% yield and 577-fold purification. The enzyme is also present in other parts of the fruit and in leaves. A MW of 12 500 was found by SDS-polyacrylamide gel electrophoresis, a value much lower that that reported for the enzyme from kidney bean fruit. Neutral or amino sugar accounts for 10% of the dry weight. In vitro, the enzyme catalysed synthesis of an unusual γ-glutamyl dipeptide which occurs in ackee seeds, using glutathione as glutamyl group donor. The enzyme mechanism was of the double displacement (ping-pong) type. View Show abstract Show more Recommended publications Discover more Article Hepatic lipid-metabolism in exercise and training May 1990 · Medicine and Science in Sports and Exercise J Gorski L B Oscai Warren K. Palmer The liver plays a central role in the metabolism of fat. The available data, though sometimes controversial, clearly indicate that muscular exercise affects almost every aspect of fat metabolism in this organ. Neither acute exercise nor training affects total lipid, phospholipid, or cholesterol concentrations in the liver of rats fed chow or low fat diets. However, exercise training reduces ... [Show full abstract] accumulation of total hepatic fat and cholesterol in rats fed a fat-rich diet. In addition, training seems to increase both the synthesis and catabolism of cholesterol in the liver in rats fed a chow diet. Production of ketones by the liver increases both during prolonged exercise and during recovery from exercise. Acute prolonged exercise reduces the activities of the enzymes involved in the synthesis of fatty acids and increases oxidation of fatty acids by the liver. This type of work also increases the esterification of fatty acids with the subsequent accumulation of triacylglycerols in this organ. Training does not affect triacylglycerol concentration in the liver of rats fed a chow diet but attenuates its accumulation after a fat-rich diet. Training reduces the postheparin plasma hepatic lipase activity. Finally, it reduces production of triacylglycerols and increases production of high density lipoprotein cholesterol by the liver. A large body of descriptive information has been published indicating that exercise has a dramatic effect upon hepatic lipid metabolism. The next step in this work is the identification of the molecular mechanisms responsible for these exercise-induced alterations. Read more Article [Effect of blood serum dialysate from healthy persons on several indices of lipid metabolism in rabb... July 1970 · Problemy Gematologii i Perelivaniia Krovi D K Virsaladze Read more Article Medium chain triglycerides in septic patients on total parenteral nutrition August 1988 · Clinical Nutrition André C. Bach M Guiraud Jean Philippe Gibault [...] P Bouletreau The object of this study was to compare the metabolic effects of a 10% long chain triglyceride (LCT) emulsion with those produced by a 10% emulsion of medium and long chain triglycerides, MCT/LCT. During 7 days, 20 septic patients received total parenteral nutrition. Daily between 9:00 and 21:00 hours, 0.14 ± 0.01 triglycerides/kg body weight/h was infused. Nine received an LCT emulsion, 11 an ... [Show full abstract] MCT/LCT emulsion. Venous blood samples were taken on each of the 7 days at 09:00 and 17:00 hours. No signs of complications attributable to the lipid infusion were observed. The plasma concentrations of phospholipids, triglycerides, free glycerol, non-esterified fatty acids, and sometimes cholesterol, rose during the lipid infusion. Blood ketone body levels did not increase. With both emulsions a day by day acumulation of phospholipids (at09:00 and 17:00 hours) in the plasma was observed. Cholesterol also accumulated but only with the LCT emulsion. The nitrogen balance and urinary excretion of creatinine and of 3-methylhistidine/creatinine ratio was lower with the MCT/LCT emulsion, which suggests less muscular catabolism in patients receiving that emulsion. Read more Article Regulation and Control in Complex, Dynamic Metabolic Systems: Experimental Application of the Top-Do... November 1996 · Journal of Theoretical Biology Stefan Krauss Patti A. Quant Metabolic control analysis has provided experimental tools and a precise language to understand and to describe regulation and control quantitatively in complex, dynamic metabolic systems. The top-down approaches of control analysis reduce and simplify the experiments required to analyse: (1) the potential (group control coefficients) of system parameters (blocks of reactions) to control system ... [Show full abstract] variables (fluxes, metabolite concentrations); (2) the sensitivity (group elasticity coefficients) of those blocks of reactions to the action of direct and/or indirect effectors and (3) the regulation of system variables, fluxes through blocks of reactions (partial response coefficients) or through the entire pathway (group response coefficients), by internal or external effectors. Although metabolic control analysis expresses regulation and control in terms of response or control coefficients, which are mathematically derived and have numerical values, the two terms are mathematically indistinguishable. In this paper, we define regulation and control as a consequence of hierarchy and not of mathematical function and analyse a conceptually simplified experimental system to demonstrate our distinction. Read more Discover the world's research Join ResearchGate to find the people and research you need to help your work. Join for free ResearchGate iOS App Get it from the App Store now. Install Keep up with your stats and more Access scientific knowledge from anywhere or Discover by subject area Recruit researchers Join for free LoginEmail Tip: Most researchers use their institutional email address as their ResearchGate login Password Forgot password? - [x] Keep me logged in Log in or Continue with Google Welcome back! Please log in. Email · HintTip: Most researchers use their institutional email address as their ResearchGate login Password Forgot password? - [x] Keep me logged in Log in or Continue with Google No account? Sign up Company About us News Careers Support Help Center Business solutions Advertising Recruiting © 2008-2025 ResearchGate GmbH. All rights reserved. Terms Privacy Copyright Imprint Consent preferences
7441
https://www.w3resource.com/python-exercises/numpy/python-numpy-stat-exercise-6.php
NumPy: Compute the weighted of a given array - w3resource Got it! This site uses cookies to deliver our services and to show you relevant ads. By using our site, you acknowledge that you have read and understood our Privacy Policy. Your use of w3resource Services, is subject to these policies More info Cookie Consent plugin for the EU cookie law  w3resource × Web Image Sort by: Relevance Relevance Date homeFront EndHTMLCSSJavaScriptHTML5Schema.orgphp.jsTwitter BootstrapResponsive Web Design tutorialZurb Foundation 3 tutorialsPure CSSHTML5 CanvasJavaScript CourseIconAngularVueJestMochaNPMYarnBack EndPHPPythonJavaNode.jsRubyC programmingPHP ComposerLaravelPHPUnitDatabaseSQL(2003 standard of ANSI)MySQLPostgreSQLSQLiteNoSQLMongoDBOracleRedisApollo GraphQLAPIGoogle Plus APIYoutube APIGoogle Maps APIFlickr APILast.fm APITwitter REST APIData InterchnageXMLJSONAjaxExercisesHTML CSS ExercisesJavaScript ExercisesjQuery ExercisesjQuery-UI ExercisesCoffeeScript ExercisesPHP ExercisesPython ExercisesC Programming ExercisesC# Sharp ExercisesJava ExercisesSQL ExercisesOracle ExercisesMySQL ExercisesSQLite ExercisesPostgreSQL ExercisesMongoDB ExercisesTwitter Bootstrap ExamplesOthersExcel TutorialsUseful toolsGoogle Docs Forms TemplatesGoogle Docs Slide PresentationsNumber ConversionsLinux TutorialsQuizzesArticles NumPy: Compute the weighted of a given array Last update on May 05 2025 11:15:51 (UTC/GMT +8 hours) Discover more Programming Statistics Library Numpy Python py numpy Weighted Average of an Array Write a NumPy program to compute the weighted of a given array. Sample Solution: Python Code: ```python Importing the NumPy library import numpy as np Creating an array 'x' using arange with 5 elements x = np.arange(5) Displaying the original array 'x' print("\nOriginal array:") print(x) Creating weights from 1 to 5 using arange weights = np.arange(1, 6) Calculating the weighted average of the array 'x' using np.average() and 'weights' r1 = np.average(x, weights=weights) Calculating the weighted average manually r2 = (x (weights / weights.sum())).sum() Asserting if the results from np.average() and manual calculation are close assert np.allclose(r1, r2) Displaying the calculated weighted average of the array 'x' print("\nWeighted average of the said array:") print(r1) ``` Copy Sample Output: Original array: [0 1 2 3 4] Weighted average of the said array: 2.6666666666666665 Explanation: In the above code – x = np.arange(5): An array x is created using the numpy.arange function to generate the values [0, 1, 2, 3, 4]. weights = np.arange(1, 6): Here numpy.arange generates the values [1, 2, 3, 4, 5], which will be used as the weights for the weighted average calculation. r1 = np.average(x, weights=weights): The numpy.average function is then used to calculate the weighted average of x using the weights array. The resulting value is assigned to r1. r2 = (x(weights/weights.sum())).sum(): This code calculates the weighted average manually using the formula r2 = (x(weights/weights.sum())).sum(). assert np.allclose(r1, r2): Finally, the code uses the numpy.allclose function to assert that r1 and r2 are equal within a tolerance. It returns true as r1 and r2 are equal. For more Practice: Solve these Related Problems: Write a function that calculates the weighted average of a 1D array given a separate weight array, ensuring normalization. Create a program that computes the weighted average along a specified axis of a 2D array using np.average with the weights parameter. Implement a solution that handles missing or zero weights by substituting them with default equal weights before averaging. Develop a method to compare the weighted average with the unweighted mean and output the percentage difference. Go to: NumPy Statistics Exercises Home ↩ NumPy Exercises Home ↩ PREV :Median of Flattened Array NEXT :Mean, Standard Deviation, and Variance Along Second Axis Python-Numpy Code Editor: Have another way to solve this solution? Contribute your code (and comments) through Disqus. What is the difficulty level of this exercise? Easy Medium Hard Based on 476 votes, average difficulty level of this exercise is Medium . Test your Programming skills with w3resource's quiz. Follow us on Facebook and Twitter for latest update. 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7442
https://jxshix.people.wm.edu/math410-problem-solving/OGF.pdf
Ordinary Generating Functions The ordinary generating function (OGF) for a series ¡a¿ of complex numbers is the formal power series P∞ n=0 anxn. When an counts the objects in a universe A for which an index parameter X has value n, we say that the generating function is indexed by X. • an be the number of binary lists of length n. Then the OGF is P∞ n=0 2nxn, which is indexed by length of binary lists. • Consider k-subsets of an n-set. Then An(x) = Pn k=0 n k  xk, is indexed by size of subsets of [n]. Given two formal power series P∞ n=0 anxn and P∞ n=0 bnxn, the sum and product (or convolution) is defined as follows, respectively: sum : ∞ X n=0 (an + bn)xn, product : ∞ X n=0 n X j=0 ajbn−j  xn. Define A(x)−1 to be B(x) such that A(x)B(x) = 1 = 1x0 + 0x1 + 0x2 + . . .. Then we may show that A(x) has a multiplication inverse if [x0]A(x) ̸= 0. Examples: (1) Multisets from two types of objects. the OGF is ∞ X k=0 (k + 1)xk = ( ∞ X k=0 xk)2. (2) Multisets (selections with repetition. let dk be the number of multisets of size k from n types of objects. Then the OGF is X k≥0 dkxk = ( X k≥0 xk)n = X k≥0 k + n −1 n −1  xk. (3) Multisets with restricted multiplicities. When S is the set of multiplicities al-lowed for a type of objects, the generating function for that factor is P k∈S xk. • When no restriction for multiplicity, then the OGF is P k≥0 xk = (1−x)−1. • For ordinary subsets, S = {0, 1}, then the OGF is (1 + x)n. • When each type must be used, the OGF is x + x2 + x3 + . . . = x(1 −x)−1. • When the usage for a type must be even, the factor for it is x2 +x4 +. . . = (1 −x2)−1. 1 In how many ways can one pick 20 coins that are pennies, nickels, or dimes, with at least three nickels and at most four dimes. Solution: the number is x20−1(x3 + x4 + . . .)(1 + x + x2 + x3 + x4). We may operate on the OGFs and on the expansions without losing equality. This may allow us to use OGF to prove something pleasantly. Some basic propositions: (1) shifting the index. X k≥0 k r  xk = X k≥r xk = X k≥0 k + r r  xk+r = xr (1 −x)r+1. Let A, B, C be are the OGFs for < a >, < b >, < c >, respectively, then (2) cn = an + bn for all n if and only if C(x) = A(x) + B(x). (3) cn = Pn i=0 aibn−i for all n iffC(x) = A(x)B(x). (4) bn = ( an−k, for n ≥k 0, for n < k if and only if B(x) = xkA(x). (5) cn = Pn i=0 ai for all i if and only if C(x) = A(x) 1−x . (6) bn = ( an for n even 0 for n odd if and only if B(x) = 0.5(A(x) + A(−x)). (7) bn = ( 0, for n even an, for n odd if and only if B(x) = 0.5(A(x) −A(−x)). (8) bn = ( ak, for n = mk 0, otherwise if and only if B(x) = A(xm). (9) bn = nan if and only if B(x) = xA′(x). Examples: • P k n k  = n2n−1. The sum is the value at x = 1 after differentiating P n k  xk. • P n 2i  . The sum is the value at x = 1 of P n 2i  x2i. • P k≥0 km. Consider P k≥0 kmxk. • Pr k=0 m k  n r−k  . The convolution of ak = m k  and bk = n k  . • Pn k=0 k(n −k). The convolution of ak = bk = k. Snake Oil method: When n is a parameter in a sum, we can always multiply by xn to form an OGF A(x). Then interchange the order of summation in the expression for A(x) and perform the new inner sum on n explicitly. (1) P k≥0 k n−k  . Let an = P k≥0 k n−k  with A(x) = P n≥0 anxn. Then A(x) = X n≥0 X k≥0  k n −k  xn = X k≥0 xk X n  k n −k  xn−k = X k≥0 xk(1 + x)k = 1 1 −x −x2. We already know that 1 1−x−x2 is the OGF for Fibonacci numbers. (2) P k m k n+k m  = P k m k n k  2k. Multiply both sides by xn and sum over n, and interchange the order of the summation. LHS = X k m k  x−k X n≥0 n + k m  xn+k = xm (1 −x)m+1 X k m k  x−k = xm(1 + x−1)m (1 −x)m+1 = (1 + x)m (1 −x)m+1. RHS = X k m k  2k X n≥0 n k  xn = 1 1 −x X k m k  2k( x 1 −x)k = 1 1 −x(1+ 2x 1 −x)m = (1 + x)m (1 −x)m+1. (3) Pn k=m c(n, k) k m  = c(n + 1, m + 1), where c(n, k) denote the number of permu-tations of [n] with k cycles. The OGF is Cn(x) = x(x+1)(x+2) . . . (x+n−1) = x(n). n X k=m c(n, k)  k m  xm = X k c(n, k) X m  k m  xm = X k c(n, k)(1 + x)k = (1 + x)(n). n X k=m c(n, k)  k m  = xm(n) = xm+1(n+1) = c(n + 1, m + 1). The generating function method to solve recursive relations • The generating function for a sequence < a > of complex numbers is the formal power series P∞ n=0 anxn or any function with power series expansion P∞ n=0 anxn. The coefficient operator [xn] extracts the coefficient of xn in a power series in x, so [xn]A(x) = an when A(x) = P∞ n=0 anxn. • Examples: Regions among lines, again an = an−1 + n for n ≥1, with a0 = 1. From an = an−1 + n, we have anxn = an−1xn + nxn, and X n≥1 anxn = X n≥1 an−1xn + X n≥1 nxn. Let A(x) = P n≥0 anxn, then P n≥1 anxn = A(x) −1 and P n≥1 an−1xn = xA(x). Note that P n≥1 nxn = x P n≥1 nxn−1 = x d dx(P n≥0 xn) = x d dx((1−x)−1). Therefore A(x)−1 = xA(x)+x(1−x)−2, and A(x) = (1−x)−1 +x(1−x)−3. We get an = [xn]A(x). • Lemma: (1 −x)−n = ∞ X k=0 k + n −1 n −1  xk. • Theorem: Let α1, . . . , αr be distinct numbers satisfying the following equation for complex numbers c1, . . . , ck with xk ̸= 0 Q(x) = 1 −c1x −c2x2 −. . . −ckxk = r Y i=1 (1 −αix)di. Then the following are equivalent for a sequence < a >. (1) < a > satisfies the recurrence an = c1an−1 + . . . + ckan−k for n ≥k. (2) < a > has generating function P(x)/Q(x) for some polynomial P of degree less than k. (3) < a > has generating function Pr i=1 Fi(x)(1 −αix)−di, where each Fi is a polynomial of degree less than di. (4) an for n ≥0 is given by the formula an = Pr i=1 Pi(n)αn i , where each Pi is a polynomial of degree less than di. • More examples: (1) Solve an = 4an−1 −5an−2 + 2an−3 for n ≥3, with (a0, a1, a2) = (2, 4, 7). (2) Solve an = 5an−1 −6an−2 + 2n2 −n + 2n for n ≥2, with a0 = a1 = 1. (3) Solve an = Pn k=1 ak−1an−k for n ≥1, with a0 = 1.
7443
https://www.metmuseum.org/art/collection/search/45673
Unidentified artist - The Four Seasons - China - Ming dynasty (1368–1644) - The Metropolitan Museum of Art Visiting Sleeping Beauties: Reawakening Fashion? You must join the virtual exhibition queue when you arrive. If capacity has been reached for the day, the queue will close early. Learn more Jump to contentticketsMember | Make a donation Search Visit Plan Your Visit Buy Tickets Become a Member Free Tours Museum Map Food and Drink Accessibility Group Visits Exhibitions and Events Exhibitions Events Free Tours Performances Art The Met Collection Curatorial Areas Conservation and Scientific Research Learn with Us Learning Resources Publications Timeline of Art History Workshops and Activities Articles, videos, and podcasts Research Libraries and Research Centers Shop Search Go The Collection The American WingAncient Near Eastern ArtArms and ArmorThe Michael C. Rockefeller WingAsian ArtThe CloistersThe Costume InstituteDrawings and PrintsEgyptian ArtEuropean PaintingsEuropean Sculpture and Decorative ArtsGreek and Roman ArtIslamic ArtRobert Lehman CollectionThe LibrariesMedieval ArtMusical InstrumentsPhotographsAntonio Ratti Textile CenterModern and Contemporary Art × Crop your artwork: Scan your QR code: Gratefully built with ACNLPatternTool The Four Seasons Unidentified artist Formerly attributed to Li TangChinese 15th century Not on view The theme of this painting is the progress of the seasons from spring to winter. The scroll begins with a scene of jagged peaks and mist-filled valleys: trees are tinged with the delicate red of new leaf buds and blossoms; turbulent mountain streams suggest the melting snow of early spring. The composition next opens into a broad vista of a lake whipped by a summer rainstorm. Autumn, with its clear skies, is a time for climbing, and the perspective shifts to high mountains with a view across distant pinnacles. In the final scene, wintry sky and water appear dark against snow-covered mountains, where only the pine and bamboo remain green. In its combination of eclectic Northern and Southern Song compositional elements and brush techniques, this landscape belongs to the revival of Song styles patronized by the conservative early Ming court. Kept alive during the fourteenth century by such artists as Tang Di (ca. 1296–1364), this style gained new prominence during the early fifteenth century in the hands of such artists as Dai Jin, Zhou Wenjing, and Li Zai. View more Listen to experts illuminate this artwork's story 7626. The Four Seasons 0:00 RW Skip backwards ten seconds. FW Skip forwards ten seconds.0:00 Your browser doesn't support HTML5 audio. Here is a link to download the audio instead. We're sorry, the transcript for this audio track is not available at this time. We're working to make it available as soon as possible. View Transcript This image cannot be enlarged, viewed at full screen, or downloaded. Public Domain Open Access As part of the Met's Open Access policy, you can freely copy, modify and distribute this image, even for commercial purposes. API Public domain data for this object can also be accessed using the Met's Open Access API. Share Link copied to clipboard Facebook Twitter Pinterest Animal Crossing Email Download image Enlarge image This artwork is meant to be viewed from right to left. Scroll left to view more. Artwork Details Use your arrow keys to navigate the tabs below, and your tab key to choose an item Overview Signatures, Inscriptions, and Markings Provenance Exhibition History References 明 佚名 舊傳李唐 四時山水圖 卷 Title:The Four Seasons Artist:Unidentified artist Chinese, active 15th century Artist: Formerly attributed to Li Tang (Chinese, ca. 1070s–ca. 1150s) Period:Ming dynasty (1368–1644) Date:15th century Culture:China Medium:Handscroll; ink and color on silk Dimensions:Image: 13 13/16 in. x 18 ft. 8 3/8 in. (35.1 x 569.9 cm) Overall with mounting: 16 5/16 in. x 35 ft. 1/4 in. (41.4 x 1067.4 cm) Classification:Paintings Credit Line:Bequest of John M. Crawford Jr., 1988 Object Number:1989.363.45 View more Inscription: Artist’s signature (1 column in standard script) Xigu, Li Tang 晞古李唐 Label strip Nagao Kō 長尾甲 （Japanese, 1864–1942）, 1 column in standard script, undated; 1 seal: 李晞古四時山水卷 老雨署。 [印]： 長尾甲印 Colophons Luo Zhenyu 羅振玉 （1866–1940）, 4 columns in semi-cursive script, dated 1937; 2 seals: 北宋畫派以荊、關及董、巨為兩大宗，董、巨一派至元而大昌，荊、關則衣缽甚廣，然至二馬以後，稍傷雄獷。此卷出天水中葉，清勁沉著，真得荊、關神髓。雖不能遽定作者姓名，然必為高手無疑。卷末李唐款為後人所加，蓋不知畫派者所為也。如此妙跡，豈必託名于晞古而後足重哉。丁丑十月上虞羅振玉觀並題記。 [印]： 羅振玉印、文學侍從 Nagao Kō 長尾甲 （Japanese, 1864–1942）, 18 columns in standard script, dated 1935; 2 seals: 洪谷畫風一變而為咸熙，再變而為晞古，以開馬、夏一派矣。按《圖繪寳鑑》：“李唐，字晞古，河陽三城人。徽宗朝，曾補入畫院，建炎間，大尉邵淵薦之，奉旨授成忠郎，畫院待詔，賜金帶，時年近八十。善畫山水、人物，筆意不凡，高宗雅愛之，嘗題《長夏江寺卷》上云：‘李唐可比李思訓，今觀此畫，骨格陗勁，頗復雄厚，能引人勝處固非常流所辦到。”上虞羅叔言則謂“卷末李唐款為後人所加，蓋不知畫派者所為也。”予竊謂題款真假，姑措不論，唯不知畫派一語未能以老友之故雷同。予審其筆，陗勁出于洪谷，雄厚得自咸熙，合二家之長鎔鑄一爐，又變其法，以闢法門，雖變於古，而不遠于古，幾似晞古以外，無能得此妙者，何可謂不知畫派乎。則定為晞古作，孰為不可，若其題款，或為後人所加，亦瑕不掩瑜，博雅所不棄焉。守屋先生好古精鑑，不知以予言為然耶？否耶？昭和十年乙亥二月雨山七十二叟長尾甲 [印]：長尾甲印、雨山 Zhang Daqian 張大千 （1899–1983）, 10 columns in semi-cursive script, dated 1951: 晞古於宣政間已享盛名，南渡後巍然畫院領袖。予嘗見其《長夏江寺》、《晉文復國》二卷，俱從長安[點去]春散出者，《江寺》一卷大青大綠，高宗題所謂可比唐李思訓者，《復國圖》人物高三、四寸，其樹石補景與此正同，蓋俱宣、政間所畫也。守屋先生高鑑以爲如何？辛卯十月蜀人張爰大千 Collectors’ seals Gu Luofu 顧洛阜 (John M. Crawford, Jr., 1913–1988) 顧洛阜 漢光閣 Unidentified 奇蛻 蹤囗囗書 囗囗書畫之印 Illegible: 2 Marking: View more John M. Crawford Jr. American, New York (until d. 1988; bequeathed to MMA) View more London. Victoria and Albert Museum. "Chinese Painting and Calligraphy from the Collection of John M. Crawford, Jr.," June 17, 1965–August 1, 1965. New York. The Metropolitan Museum of Art. "Great Waves: Chinese Themes in the Arts of Korea and Japan I," March 1–September 21, 2003. New York. The Metropolitan Museum of Art. "Art of the Brush: Chinese Painting and Calligraphy," March 12–August 14, 2005. New York. The Metropolitan Museum of Art. "The Four Seasons," January 28–August 13, 2006. New York. The Metropolitan Museum of Art. "Arts of the Ming Dynasty: China's Age of Brilliance," January 23–September 13, 2009. New York. The Metropolitan Museum of Art. "Streams and Mountains without End: Landscape Traditions of China," August 26, 2017–January 6, 2019. View more Weng, Wan-go, and Thomas Lawton. Chinese Painting and Calligraphy: A Pictorial Survey: 69 Fine Examples from the John Crawford, Jr. Collection. New York: Dover Publications, 1978, cat. no. 29. Suzuki Kei 鈴木敬, ed. Chûgoku kaiga sogo zuroku: Daiikan, Amerika-Kanada Hen 中國繪畫總合圖錄: 第一卷 アメリカ - カナダ 編 (Comprehensive illustrated catalog of Chinese paintings: vol. 1 American and Canadian collections) Tokyo: University of Tokyo Press, 1982, pp. 102–103, cat. no. A15-036. Shih Shou-ch'ien, Maxwell K. Hearn, and Alfreda Murck. The John M. Crawford, Jr., Collection of Chinese Calligraphy and Painting in the Metropolitan Museum of Art: Checklist. Exh. cat. New York: The Metropolitan Museum of Art, 1984, p. 26, cat. no. 53. View more Learn more about this artwork Asian Art at The Met The Met's collection of Asian art—more than 35,000 objects, ranging in date from the third millennium B.C. to the twenty-first century—is one of the largest and most comprehensive in the world. Timeline of Art History Chronology Central and North Asia, 1400-1600 A.D. Chronology China, 1400-1600 A.D. Related Artworks All Related Artworks By Li Tang By Unidentified artist Asian Art Handscrolls Ink Paintings Scroll paintings Silk From Asia From China From A.D. 1400–1600 Portrait of a Member of the Family (?) of Emperor Tongzhi (1862–74) Unidentified artist Chinese, 19th century 19th century Mountains in the Snow Unidentified artist Illustrated Story of Yoshitsune Unidentified artist One Hundred Poems of Moritake Illustrated with Comments Unidentified artist Illustrated book Unidentified artist Resources for Research The Met's Libraries and Research Centers provide unparalleled resources for research and welcome an international community of students and scholars. The Met Collection API is where all makers, creators, researchers, and dreamers can connect to the most up-to-date data and public domain images for The Met collection. Open Access data and public domain images are available for unrestricted commercial and noncommercial use without permission or fee. Feedback We continue to research and examine historical and cultural context for objects in The Met collection. If you have comments or questions about this object record, please complete and submit this form. The Museum looks forward to receiving your comments. The Met Fifth Avenue 1000 Fifth Avenue New York, NY 10028 Phone: 212-535-7710 The Met Cloisters 99 Margaret Corbin Drive Fort Tryon Park New York, NY 10040 Phone: 212-923-3700 About The Met Mission and History Collection Areas Conservation Departments Accessibility Press Support Membership Host an Event Travel with The Met Corporate Support Career Opportunities Volunteers Fellowships Internships Follow us Join our newsletter Sign Up Accessibility Site Index Terms and Conditions Privacy Policy Contact Information © 2000–2024 The Metropolitan Museum of Art. All rights reserved.
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https://www.youtube.com/watch?v=72TJ48vnvqo
Teaching Ratio for Conceptual Understanding Using Models and Concrete Manipulatives Shelley Gray 4900 subscribers 6 likes Description 1099 views Posted: 19 Mar 2024 Today we are taking a look at how ratio can be made conceptual through the use of concrete manipulatives and model drawing. I love this, because it takes a concept that has the potential to be quite abstract and procedural to conceptual and simple to visualize. Get more support for using models and manipulatives in your classroom here: teachingratios #modeldrawing #math #mathteacher #mathteachersunite #iteachmath #iteach456 #problemssolving #mathproblems Transcript: hey there and welcome to the next video in this series this one is all about ratio I'm pretty excited about this one because ratio is often taught very procedurally um like a series of steps but when we incorporate models and manipulatives we can give kids a real real conceptual and deeper understanding of what ratio really means so we are going to start out with some quiz and air rods you could ask your students how they might represent this problem using the rods I guarantee you that you'll have some some kids take the red rods and the blue rods and use them to represent the groups of cars this is a really good opportunity though to discuss how our parts need to all be the same size so that won't work now after you've had a really good discussion with concrete manipulatives we can relate that to a bar model as I'm showing here so we know that our ratio is 2 to 3 all of our parts are equal sizes we know that there are 15 blue cars so we can simply divide that by three parts to figure out that there are five cars in each part which means that there are also 10 red cars so our total is 25 so as we can see from the example that we just did using models like this for a concept like ratio gives our kids the opportunity to develop that conceptual understanding when they can actually visualize what it means it's a lot easier to understand okay let's take a look at another problem this one's a little bit different this time we have a ratio of zebras to giraffes which is 3 to four and we know that there's 56 animals in all this time we know the total so we're going to draw our bar model and we can see that all together there are seven parts and there are 56 animals so what we can do is simply divide 56 by seven parts and that tells us that there are eight animals in each part now it's really simple to see how many of each type of animal there are at the zoo
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https://www.youtube.com/watch?v=W2kVG9qbYK8
19 maximum and minimum distance of a point from circle by kota faculty H.D. MATHEMATICS 116000 subscribers 39 likes Description 736 views Posted: 20 Jul 2020 maximum and minimum distance of point form circle by Hariom dubey Hd sir ex faculty of resonance kota 2 comments Transcript: कि भारत माता की जय भारत माता की जय तक बच्चों हम लोग सर्कल के अगले पार्ट पर आ गए हैं और इस पाठ में हम लोग एक और बढ़िया कौन से पढेंगे सर्कल का जो यह कांसेप्ट क्या है यह कौन से फै मैक्सिमम कि एंड मिनिमम ए स्टैंड स्टिल कि आफ ए प्वाइंट फ्रॉम सरकार को मैक्सिमम एंड मिनिमम डिस्टेंस आफ ए प्वाइंट फ्रॉम थम है तो जैसे हम लोग आगे चलकर ट्रैवल्स में मैक्सिमा एंड मिनीमा पड़ते हैं है तो वहां पर भी इस कांसेप्ट का कई बार एप्लिकेशन हो जाता है काफी अच्छा है है तो क्या यह कौन से स्टेट समझो जैसे कि मान लो अगर मेरे पास एक सर्कल है और एक पॉइंट है जो सरकार के बाहर है अब सेंड ए कि हमने क्या समझाया था पॉइंट सर्किल के बाहर का होता है जब S1 पोस्टिव होता है एस वन का मतलब सरकार क्वेश्चन में एक सिपाही की जगह पर यह पॉइंट रख दो अरे ऐसे मन कितनी बार है इसमें तो भूल गए क्या अच्छा-अच्छा याद है ना इसमें क्या होता है तरक्की क्वेश्चन में एक सिपाही की लेकर पर इस पॉइंट को तो यह स्पष्ट हो जाता है समझ गया ये तो ईश्वर नगर पॉइंट पौष्टिक होता है तो पॉइंट सर्कल के बाहर होता है अच्छा अब देखो अगर हम एक ऐसी लाइनों करें मैं यहां से जो सरकार के सेंटर से पासबुक में एक ऐसी लाइनों करें इस प्रकार के सेंटर से पास को तो को यहां से यह छुप्पी के डिस्टेंस जपिए यह हो गई पॉइंट की सर्कल्स मिनिमम डिस्टेंस मैं इतनी मिनिमम और मैक्सिमम डिस्टेंस होना यह हमेशा नार्मल के इलाकों होती है मैं पागलों के भी हम लोग बात देखेंगे ना प्राकट्य जब मैं जब मैं पढेंगे तो वहां पर भी कौन सा लगता है है एसिड यह नो क्या हो गई यह लाइन सरकार नॉर्मल हो गई नार्मल हमेशा सेंटर से पास होता है सर कल का शो है तो मैक्सिमम और मिनिमम डिस्टेंस जो होती है वह हमेशा सर्कल के या किसी भी कोण कि नॉर्मल किला होती है अच्छा ठीक है तो पीछे डिस्टेंस है यह हो गया मिनिमम डिस्टेंस को मिनिमम डिस्टेंस आफ पी थे प्रॉस वर्क है और जो यह डिस्टेंस हो जाएगा सीटी - ऑफ ए आर के हो गया 8 कोई सरकार करेगी फैसला कि यह आरएस करे लिए बहुत अच्छी टीम लोकायुक्त का सेंटर तो मालूम हुआ कि यह सरकार दिया जाएगा मतलब इसका सेंटर दिया गया है और पॉइंट भी दिया जाएगा तो सीधी निकल आएगा करे डिस्टेंस बिटवीन टू प्वाइंट्स यार सीमा दो मैं पी मामलों में डिस्टेंस बिटवीन टू प्वाइंट्स भाई में रहे और सड़कों पर करीब मालूम होगा है तो सीवरेज मालूम होगा और आर्मी मालूम होगा तो हम मिनिमम डिस्टेंस को निकाल लेंगे है लेकिन यहां पर हम यह मानकर चल रहे पॉइंट सर्कल के बाहर है ध्यान देना अजय को कि यह जो पीबी है ना पीवी यह कई बार कंपलेक्स नंबर में बीए 570 कंपलेक्स नंबर के और कोऑर्डिनेटर के बहुत से कौन से मिलते हैं कांग्रेस की जिम्मेदारी होती ना उसका कांटेस्ट से बहुत गहरा रिश्ता है गहरा दोनों संबंधी है कि Bigg Boss था किसका पॉलिश कुंठित कर दो है तो कोऑर्डिनेटर परदे से बड़े परदे होती बच्चों को मैं इसे ऐसे ही पैसा टॉपिक मत समझना कि बहुत ही इंपोर्टेंट टॉपिक है देखो पीरी पीरी क्या हो गया MP3 हो गया मैक्सिमम डिस्टेंस फोटो कॉपी फ्रॉम सर्किल स्तर कल से इस पॉइंट अमेज़न डिस्टेंस और तो यह हो जाएगा सीपी प्लस आ कुछ यादें करना ही कॉमर्स साइंस सिटी प्लस पार्थ कि मैं सी मालूम होगा और पी मालूम होगा सिंह तोमर लकी बंद होगा सर्कल गिवर सेंटर निकल है ना पॉइंट टीके मनोभाव c14 केवल होगा तो सिंपली डिस्टेंस फ्रॉम गिवन होगा सर्कल के वेंस का रेडियस भी निकल जाएगा पहुंच गए कि यह पहन सकता है जब पॉइंट सरकार के बाहर तो अगर यह पार ने 2010 से पर कुछ नहीं पा मुझे बहुत कम हर चीज है कॉमन सेंस एस में समझ दिमाग में बैठा हूं कि टिकट सीटी मजेदार - कर देंगे सीटी में जोड़ देंगे लेकिन तब जब पॉइंट बाहर हो यह कह सकता है बट आज अपॉइंट बाहर हों तो पहले पहचानना पड़ेगा पॉइंट बाहर है कि नहीं पॉइंट बाहर का होता है जब से संबंधित पोस्ट होता है उन्होंने इस बात को कुछ लिखो जल्दी लिखो फटाफट आ जी हां ये दुनिया में तेरा है बड़ा नाम आज मुझे भी तुझसे पड़ गया पांव मेरी विनती सुने तो जानू मानू तुझे मैं राम राम नहीं तो सारी दुनिया में तुझको तू ध्यान दूं अगर पॉइंट आपके अंदर हूं की मांग सरकार जब यह सेंटर यहां पर पॉइंट है सर्कल के अंदर तो पॉइंट सर्कल के अंदर का होता है जब फेस्टिवल निगेटिव होता है यह टिप्स S1 धो लें तो हरियाणा सर्कल का क्वेश्चन लो उसमें एक्स राय की जगह इस पी के कोऑर्डिनेटर को रख दूं तो क्या हो जाता है ऐसे वक्त अच्छा ठीक है टैक्स की जगह एक्सप्रेस-वे की जगह भाई मस्त अच्छा ठीक है एस वस्त एस मतलब डिस्टिक कोऑर्डिनेटर है क्या एक्स 1.1 है मैंने मान लीजिए तो इसको बोलते हैं इसमें ही मारुति के प्वाइंट्स 151 ओ मा लिए केवल बाइबिल और मालूम सरकार लिए हैं एक्स स्क्वायर प्लस सब्सक्राइब तो यह कहते हैं तो स्वप्न दोष नहीं होगा इसको वाले सब्सक्राइब की जड़ दिया है कि अ कि इस बार निगेटिव होगा तक पॉइंट सर्कल के इनसाइड होगा ज्ञान तो कि अब जैसे हम एक लाइन रोग है जो नार्मल के अलार्म को नार्मल करेंगे तो यह पॉइंट मान लो यह है पर यह भी है है तू जो यह हमारा सीधा है ना है अब यह जो पीछे होगा ना यह पॉइंट का सर्कल से मिनिमम डिस्टेंस तो यह हो गया मिनिमम डिस्टेंस में अपील फ्रॉम सरकार अपने आप लेकिन यार बार-बार वही प्यारी हैं पॉइंट का डिस्टेंस दिखा दो पूरा डिस्टेंस का पॉइंट सर्कल से मिनिमम और मैं इसमें दो इस तिलक पिए निकालोगे कैसे दुख ही अपना रेडियो से क्या बोलो आ कि यह आदमी तो आम मैसेज सीटी - करोगे सीपी निकल आएगा सेंटर वालों में पॉइंट मनोस्थिति आ जाएगा आर्मी मालूम होगा तो यह हो गया मिनिमम डिस्टेंस चेक आपको यहां से जो पीड़ित होगा पी बी पी बी यह होगा क्या बोलो मैक्सिमम डिस्टेंस है और पीली क्या होगा बोलो तो और पीवी पीस तक यह भी आरोप है कि यदि आम हो गया इस सेंटर तो यह हो जाएगा पति प्लस एप में पीली का मतलब सीपी डिस्टेंस प्लस अा ग्य हो जाएगा पॉइंट का सर्कल से मैक्सिमम डिस्टेंस अच्छा ठीक है कि मैं बता दे रहा हूं यह पारूले मान के याद मत करने लग जाना इतनी की याद नहीं कर सकते हो आप भूल जाओगे सब कुछ भी कॉमन सेंस है कि इस राहुल को कि अगर पॉइंट सरकार के ऊपर ही आ जाए मालूम पॉइंट यहां पर आ गया आती सर्कल के ऊपर यह आ गया भी आप MS Word क्या हो जाएगा 0 कि ऐश्वर्या अब क्या हो जाएगा बोलोगी अब यह पॉइंट चैनल को सब्सक्राइब करने का स्पंदित हो जाएगा तो कि यह सेंट्रल यह नॉर्मल आप तो खूब पॉइंट की सरकार से मिनिमम डिस्टेंस जीरो हो गई बोलो आप क्या हो गई मिनिमम जो डिस्टेंस हो गई ना यह पॉइंट की तरह से मिनिमम डिस्टेंस हो गई थी वह और जो मैक्सिमम डिस्टेंस हो गई राम मैक्सिमम डिस्टेंस कि यह हो गई टीवी के प्रति भी काम अंदर कितना हो गया X2 आ चुकी थी तो यह भी आ और यह भी आंख तो पी बी मजूमदार डिस्टेंस कितना हो गया ब्लुटूथ है हैं पर मिनिमम डिस्टेंस कॉफी मिनिमम का मतलब सर्कल से क्वांटिटी मिनिमम डिस्टेंस वह जीरो हो गई पर यहां पर मिनिमम डिस्टेंस तो यह हो गई और मैग्नीशियम डिस्टेंस यह बिलकुल अ तो पॉइंट कहां है वो यह देखकर ही अगला स्टेप लिया जाएगा और यह कोई फार्मूले नहीं है यह सब कांसेप्ट है देखकर खुद समझ जाओगे क्या चल रहा है यह उसको ठीक है तो तीन केस बने पॉइंट बाहर है पॉइंट अंदर है है और पॉइंट सर्कल के ऊपर एक पिस्टल और बना दो पॉइंट उसका सेंटर ही है कि एक हिस्सा और बना सकते हो व्यक्ति का की पॉइंट उस सर्कल के सेंटर के ऊपर ले लिया है यह पॉइंट है मुझसे इंटर के ऊपर ले लिया कि अब देखो यह नार्मल ड्रॉप करोगे तो मिनिमम डिस्टेंस भी रेडियस के बराबर और मैक्सिमम डिस्टेंट रिलेटिव्स के ही बराबर पॉइंट का सर्कल से यह डिस्टेंस यही मिनिमम हो गया यह मैक्सिमम हो गया है ना तथा पॉइंट माल जो सेंटर ही हो गया तब है तो मिनिमम पर यह हो गया और मैं इसमें भी क्या हो गया आ रही हो गया मिनिमम भी और यहां पर इस केस में मिनिमम भी आज और मैक्सिमम में कितना हो गया आप अगर पॉइंट सेंटर ही हो गया है पर ऐसे में सीटी की हो जाएगा तो यह पॉइंट सेंटर के ऊपर ही चला गया तो सेंटर प्वाइंट के बीच का डिस्टेंस जीरो हो जाएगा ना ऐसी यहां पर सीटी मिलेगा एसीपी की हो जाएगा आप समझ गए ना हां अब sms निगेटिव व क्योंकि पॉइंट अभी भी सर्कल के अंदर यहां पर ऐसे माना भी नेगेटिव होगा पर सीपी हो जाएगा जागरूक है कि जो सेंटर है वह पॉइंट 800 तो हो नहीं होता है क्वेश्चन करता हूं उसका वीडियो पर जल्दी से फटाफट पास टेक्नो कि यशोमती मैया से बोले नंदलाला रे कि राधा क्यों गोरी मैं क्यों काला कि राधा क्यों गोरी मैं क्यों काला वो बोली मुस्काती मैया सुन मेरे ना कि तुम क्वेश्चन कर लिया गया है थे फाइंड मिनिमम लिए बेड मैक्सिमम distance of your 361 हाउ टू पर तन को कि एक्स स्क्वायर प्लस बी स्क्वायर प्लस टू एक्स माइनस टू माय - अिववािहत एक पॉइंट का सरकार से मिनिमम एग्जाम डिस्टेंस है तो सब काम क्या होगा इस वक्त ए एस वन निकालते हैं मालूम चलेगा आकर यह पॉइंट अंदर व्यवहार है तभी तो अभी कैलकुलेशन होगी महाराज SMS निकालो कैसे एक्सेस की जगह चाहिए आपको अरे भाई की जगह फिक्स रखो तो एक टैक्स की जगह ट्री ओं कि इसकी जगह अच्छी तो और भाई की जगह फिक्स चीज 6 क्या हुआ बोलो नारायण प्लस फटी 6 प्लस यॉर्क पोस्ट इच्छुक और सीरियल देखना है जो ऐश्वर्या यह पॉजिटिव है मेरे भाई क्लियर देख रहा है है तो पॉइंट बहाने कि अगर पॉइंट अपना बाहर के बच्चों यह सेंटर ने यह पॉइंट है यह लाइन ड्रा करेंगे इन्होंने कि नॉर्मल तो पिए क्या होगा मिनिमम टीवी क्या होगा मिक्स इट इज इक्वल टू कि इस प्रकार का सेंटर क्या हो गया बोलो ना है तो इसमें इस सर्कुलर में 2जी X2 ऊंची बराबर 3G हो गया 3 बिट्टू है क्या होता है एक्सेस कर बिस्किट 2G एक्स और टू फटी तो एफिल टावर - टू पसीना हो गया - बंद तो अफ्रीका हो गया तो इस सरकार से क्या हो गया है 8 - जी मरने से तो - 352 और वक्त को पॉज रगड़कर रेडियस क्या हो गया अनिरुद्ध इस पर ए प्लस बी स्क्वायर और - सी तो प्लस वन बीएचके रुष व सब्सक्राइब - सी तो हो गया अंडर रूट 9 4 प्लस टू है तो 17.the इंफाल अच्छा ठीक है जी हां अब देखिए यहां से को मिनिमम डिस्टेंस को सीपी निकालो सी पी पी पांडेय 180 यह आप 9th सीपी निकाल लो डिस्टेंस बिटवीन टू पॉइंट गिफ्ट पुलिस ने स्टेटमेंट टू फ्रंट यह पीस है यह सी है तो क्या होगा X2 - सेक्स संबंध अच्छा ठीक है X2 - सफल का पुलिस पर और Y2 - व्वा पुष्कर यह हो गया हमारा सिटी सीओ पीके रिकॉर्ड यह पॉइंट एस ई ई मैं सिर्फ तो क्या हो गया और नो नो बट ए तू तो 814 प्लस प्रैक्टिस हां तो क्या हो गया 181 बट यह हो गया हमारा जो मिनिमम डिस्टेंस हो गया को मिनिमम डिस्टेंस डिस्टेंस सीटी - 5 कि यह वह क्या चीज पे ए पी पी और जो यह है कितना बोलो आप यह भी आप यह कितना हो गया सिटी वाइस सिटी मामलों में मैं क्या बोलू अंडर रूट 181 भारतेंदु 4 को बाहर गए रुकी तो आ जाएगा यह दोनों कार इसमें भी चार को पहन लेंगे तो द्वारा जैसी मनासा हैं और अगर हमसे पूछा जाए बच्चों क्या हक मैक्सिमम डिस्टेंस अब वह भी तुम्हारे बाएं हाथ का खेल है भाई तो यह हो जाएगा टीवी पर देख रहा है मस्तिष्क मैक्रम डिस्टेंस कितना हो गया पीवीटी प्रशांत सीपी प्लस पार्ट 2 सीटी 181 बट ई दो और सब्सक्राइब बटन तू तो यह हो गया मैक्सिमम डिस्टेंस डिस्टेंस पर बहुत इंपोर्टेंट है के दूसरे मैच में है जो आईटी ने पूछा कि आखिर बताऊंगा इसमें कोई चार्ज आपको नहीं कई बार क्या होता लोग यह समझाने के लिए कि बच्चे ध्यान से पढे नोन इंपॉर्टेंट चीजों को भी इंपोर्टेंट बता देते हैं वह बच्चे के लिए सॉफ्ट होती हानिकारक होती है से एक हफ्ते के लिए यह तक हानिकारक है समझे है तो ऐसा मैं नहीं करूंगा जो इंपोर्टेंट है वह बताऊंगा इंपोर्टेंट है जो नहीं इंपोर्टेंट है कहूंगा भैया इससे बचकर निकलना चाह तो निकल लोकल टेंशन नहीं है ऐसे लिमिटिंग पॉइंट मैंने पढ़ा है मैंने कहा इंपोर्टेंट नहीं है चाहे तो मत पढ़ो और अगर बहुत बड़े जिज्ञासु हो तो पहलू जिज्ञासु रूपणी कोई दिक्कत टेंशन थोड़ी है इस दिन को हरि ओम बाबा प्रेशर को पढ़े लिखो जल्दी से पड़ा है ठीक है तो इस वीडियो में इतना ही करते हैं अगले वीडियो में मिलेंगे फिर मैंने कौन से टकराए थे
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https://www.youtube.com/watch?v=y3lYswA_jkI
Simplify Cube Roots with Exponents & Variables | Fractions | Eat Pi Eat Pi 18000 subscribers 26 likes Description 2522 views Posted: 5 Oct 2022 In this video, I teach you how to simplify cube roots that have exponents and variables. I go over examples with whole numbers, fractions, and negative numbers. There are a couple playlists attached at the end of the video if you need help with any other topics. If you have any questions, please leave them in the comment section below! Also, if you find the videos helpful, please like, share, and subscribe! 2 comments Transcript: what's up you freaking Geniuses so in this video I'm going to teach you how to simplify cube roots all right and you should be familiar with some perfect cubes so these are a few of them so 8 27 64 125 and there's a bunch more okay so these are just numbers that we can easily take the cube root of right so for example the cube root right the cube root of eight is equal to two the cube root of 27 is equal to 3 the cube root of 64 is 4. 125 is 5 right and Etc okay or you can think about it the other way right so 2 cubed would be equal to eight three cubed is equal to 27 4 cubed is equal to 64 and Etc okay so here uh with this first problem we have the cube root of 128. okay so here you just want to think of what two numbers can I multiply together to get 128 and there's a bunch of different ways we could do it but here you would specifically want to use something like 64 times two okay and the reason you'd want to do that is because because 64 is a perfect Cube and you want to use perfect cubes because it's going to simplify your math so the cube root of 128 we can break down into the cube root of 64 times the cube root of 2. okay now again 64 is a perfect cubed right it's equal to 4. so the cube root of 64 is equal to 4 and then here the cube root of 2 we can't simplify that anymore so we would just multiply this by the cube root of 2. okay so then your final answer right here would be 4 cube root of 2. okay the next one over here has some variables in it so it's cube root of 125x squared 125 just happens to be a perfect Cube so when you have a number and variables you want to break those into two separate cube roots all right so this is going to be equal to the cube root of 125 times the cube root of x squared okay and again we said this is a perfect cube right this is just equal to 5. so then here we really have 5 times the cube root of x squared now the cube root of x squared we can't break that down anymore either and the reason for that is because the exponent over here has to be a 3 or bigger in order to simplify a cube root so 2 is obviously less than three so that's why we can't break this down anymore all right so then this would be your final answer right there so 5 cube root of x squared all right so now let's do some examples where we're going to be able to break down the exponent more so here we have the cube root of 24 x cubed right so again we can split this up into the number and the variable okay so this is going to be equal to the cube root of 54 times the cube root of x cubed okay so now 54 that's not a perfect Cube but we can simplify this into 27 times 2. right and you'd want to specifically choose this pair right here because we want to try and use perfect cubes right so 27 is a perfect Cube so then this one over here this is going to be equal to the cube root of 27 times the cube root of 2 right cube root of 27 times the cube root of 2. and then here we have the cube root of x cubed and this is simply equal to X okay so then we're multiplying by X okay so let me clarify that so if I took the square root of let's say x squared this would be equal to X if I took the cube root of x cubed this would be equal to just X if I took the fourth root of x to the fourth this would be equal to just X okay and the fifth root of x to the fifth would just be equal to X okay so that's the pattern that's kind of going on here so again the cube root of x cubed is equal to just X right so we just have an X right there so to simplify this again the cube root of 27 is equal to 3 right that's a perfect cubed so this is equal to 3 times the cube root of two times x okay those are all as simplified as we can get them so that would be your final answer okay and just a couple more here so here's the negative cube root of 32x to the seventh okay so again we're going to split up the number and the variable so the cube root of 32 times the cube root of x to the seventh okay now 32 this is not a perfect Cube but we can break it down using a perfect Cube specifically a 8 right so then we can break this down into eight times four right so the cube root of eight times the cube root of four so let's write that out the cube root of eight times the cube root of four okay now we have the cube root of x to the seventh right so one way we can break down this x to the seventh is by multiplying it out as X cubed times x cubed times x right so here we have seven x's and now here we have seven X's right because three plus three plus basically one is equal to seven and the reason I want to use x cubes as much as I can is because I know the cube root of x cubed is equal to just X right so that kind of simplifies my math if I specifically break it down with an X cubed okay so that's what we're going to do we're going to break it out into the cube root of x cubed times the cube root of x cubed times the cube root of just X so we're going to multiply Again by the cube root of x cubed times the cube root of x cubed times the cube root of just X okay so now we can simplify a bunch of stuff here right so first of all cube root of 8 that's a perfect cubed that's equal to 2 and then the cube root of four that's not a perfect Cube so that just stays as the cube root of 4 and then here we have the cube root of x cubed which is equal to X same thing here times x times here is just the cube root of x right the cube root of x so if we multiply all this together we're going to get let's see first of all the numbers 2 times the cube root of four two times the cubed root of 4 times x times x so that's equal to x squared right so times x squared times the cube root of x right so times the cube root of x okay and don't forget we actually had a negative sign at the very beginning right so this means our answer is also negative okay so then our final answer right here would be negative 2 cube root 4 x squared cube root X ugly answer I know but that's as simplified as we can get all right so now let's finish up with this one so it's the cube root of eight x to the fourth over 27y cubed so we have a fraction here right so one thing you can do is actually break up the radical so we can write it as the cube root of the top so 8 x to the fourth over the cube root of the bottom so 27 y cubed okay so now we can kind of simplify the top on its own and simplify the bottom on its own so on top the cube root of 8 x to the fourth again we can split that up into the number and the variables right so the cube root of eight times the cube root of x to the fourth and on the bottom same thing splitting up the numbers and variables the cube root of 27 times the cube root of Y cubed and this is going to be equal to on top the cube root of eight that's the perfect Cube that's equal to two and then here we have the cube root of x to the fourth so this x to the fourth I'm going to break down into X cubed times x again I want to try and get one of these X cubes like this right because I know what the cube root of x cubed is so I'm going to split it up like that so we're going to say the cube root of x cubed times the cube root of x and that's all over the bottom over here cube root of 27 is equal to 3. and then that's times the cube root of Y cubed in this case which is equal to just y right just y so then simplifying this again on top we have the cube root of x cubed so that's equal to just X and then here the cube root of x is just cube root of x right and then we have our 2. so on top we're going to have 2 times x times the cube root of X all over 3 times Y which is 3y okay that's as simplified as we can get that so this would be your final answer so if you found the video helpful definitely leave a thumbs up down below and if you have any other questions or want to see any other examples just let me know in the comment section below
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1 / 63 计量经济学(Econometrics) 计量经济学(Econometrics) 胡华平 胡华平 西北农林科技大学 西北农林科技大学 经济管理学院数量经济教研室 经济管理学院数量经济教研室 huhuaping01@hotmail.com huhuaping01@hotmail.com 2023-02-15 2023-02-15 2 / 63 5.1 过原点回归 5.2 尺度与测量单位 5.3 标准化变量回归 5.4 对数线性模型 5.5 半对数模型 5.6 倒数模型 5.7 函数模型的选择 第5章：一元回归：模型形式扩展 第5章：一元回归：模型形式扩展 3 / 63 5.1 过原点回归 5.1 过原点回归 huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 4 / 63 过原点回归（regression through the origin）：没有截距项的线性模型 在实践中，双变量PRM过原点回归采取如下的形式： Yi = β2Xi + ui 适用于这种模型的例子： 弗里德曼的持久收入假说(permanent income hypothesis)； 资本资产定价模型( the capital Asset Pricing Model, CAPM )等。 过原点回归的模型形式 过原点回归的模型形式 huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 5 / 63 资本资产定价模型( the capital Asset Pricing Model, CAPM )： (ERi −rf) = βi (ERm −rf) 其中： 证券 的期望回报率； ERi i 市场证券组合的期望回报率（如标准普尔S&P500综合股票指数）； ERm 为无风险回报率（90天国债回报率）。 rf 为系数，表明第 种证券回报率与市场互动程度的度量。(注:不要把这个 和双变 量回归的斜率系数 混同起来。) βi i βi β2 一个大于1的 意味着证券 是一种易波动或进攻型证券；一个小于1的 意味着 证券 是一种防御型证券。 βi i βi i 资本资产定价模型（CAPM） 资本资产定价模型（CAPM） huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 6 / 63 Ri −rf = βi (Rm −rf) + ui Ri −rf = αi + βi (Rm −rf) + ui 如果CAPM成立，则预期 为0。 αi 这样的模型如何估计呢？ 资本资产定价模型（CAPM） 资本资产定价模型（CAPM） huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 7 / 63 这类模型的SRM可以写成： Yi = ^ β2Xi + ei OLS方法下求解回归系数： ∑e2 i = ∑(Yi −^ β2Xi) 2 = 2 ∑(Yi −^ β2Xi) (−Xi) = 0 ^ β2 = = = β2 + E (^ β2) = β2 ∂∑e2 i ∂^ β2 ∑XiYi ∑X2 i ∑Xi (β2Xi + ui) ∑X2 i ∑Xiui ∑X2 i 资本资产定价模型（CAPM） 资本资产定价模型（CAPM） huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 8 / 63 OLS方法下求解得到的方差： V ar (^ β2) = E(^ β2 −β2) 2 = E[ ] 2 = ∑Xiui ∑X2 i σ2 ∑X2 i ^ σ2 = ; E (^ σ2) = σ2 ∑e2 i n −1 资本资产定价模型（CAPM） 资本资产定价模型（CAPM） huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 9 / 63 OLS估计量对比：无截距和有截距的差异： Yi = ^ β2Xi + ei ^ β2 = V ar (^ β2) = ^ σ2 = ∑XiYi ∑X2 i σ2 ∑X2 i ∑e2 i n −1 Yi = ^ β1 + ^ β2Xi + ei ^ β2 = V ar (^ β2) = ^ σ2 = ∑xiyi ∑x2 i σ2 ∑x2 i ∑e2 i n −2 第一，对有截距项的模型来说，总有 ；对无截距项的模型来说， 不一定成立，只有 成立。 ∑ei = 0 ∑ei = 0 ∑eiXi = 0 第二，对有截距项的模型，判定系数 ；但是，对无截距模型来说， 时可能 出现负值。 r2 ≥0 r2 资本资产定价模型（CAPM） 资本资产定价模型（CAPM） huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 10 / 63 过原点回归的判定系数 的计算公式如下： r2 TSS = ∑y2 i = ∑Y 2 i −n¯ ¯ ¯ ¯ Y 2 RSS = ∑e2 i = ∑Y 2 i −^ β 2 2 ∑X2 i r2 = 1 − > 0; r2 = 1 − < 0; RSS TSS RSS TSS 因此，对于无截距模型，我们给出拟合优度指标为毛判定系数（Raw ）： r2 Raw r2 = ∑(XiYi)2 ∑X2 i ∑Y 2 i 资本资产定价模型（CAPM） 资本资产定价模型（CAPM） huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 11 / 63 启示： 第一，尽管模型含有截距项，但若该项的出现是统计上不显著的(即统计上等 于零) ，则从任何实际方面考虑，都可认为这个结果是一个过原点回归模 型。 第二，如果在模型中确实有截距，而我们却执意拟合一个过原点回归，我们 就犯了设定错误。 资本资产定价模型（CAPM） 资本资产定价模型（CAPM） huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 Showing 1 to 8 of 240 entries Previous 1 2 3 4 5 … 30 Next year month X Y 1980 1 7.2634 6.0802 1980 2 6.3399 -0.9242 1980 3 -9.2852 -3.2862 1980 4 0.7933 5.2120 1980 5 -2.9024 -16.1642 1980 6 8.6132 -1.0547 1980 7 3.9821 11.1724 1980 8 -1.1502 -11.0633 12 / 63 1980.01-1999.12年间104 种股票构成的一个指数的超额回报率 (%)和英国总体 股票指数的超额回报率 (%)的月度数据共n=240个月观测。其中超额回报率指 的是超过无风险资产回报率的部分。 Yt Xt 资本资产定价模型（CAPM）：数据 资本资产定价模型（CAPM）：数据 huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 13 / 63 下面先直接给出二者的散点图： 资本资产定价模型（CAPM）：散点图 资本资产定价模型（CAPM）：散点图 huhuaping@ 第5章：一元回归：模型形式扩展 5.1 过原点回归 14 / 63 两类模型回归结果对比： 无截距模型： Yi = ^ β2Xi + ei ˆ Y = + 1.16X (t) (15.5320) (se) (0.0744) (fitness)R2 = 0.5023; ¯ R2 = 0.5003 F ∗= 241.24;p = 0.0000 有截距模型： Yi = ^ β1 + ^ β2Xi + ei ˆ Y = −0.45 + 1.17X (t) (−1.2329) (15.5350) (se) (0.3629) (0.0754) (fitness)R2 = 0.5035; ¯ R2 = 0.5014 F ∗= 241.34;p = 0.0000 资本资产定价模型（CAPM）：回归结果 资本资产定价模型（CAPM）：回归结果 15 / 63 5.2 尺度与测量单位 5.2 尺度与测量单位 huhuaping@ 第5章：一元回归：模型形式扩展 5.2 尺度与测量单位 Showing 1 to 5 of 16 entries Previous 1 2 3 4 Next Year GPDIB GPDIM GDPB GDPM GPDIB_std GDPB_std 1990 886.6 886600 7112.5 7112500 -1.2942 -1.3459 1991 829.1 829100 7100.5 7100500 -1.4624 -1.3550 1992 878.3 878300 7336.6 7336600 -1.3185 -1.1768 1993 953.5 953500 7532.7 7532700 -1.0986 -1.0287 1994 1042.3 1042300 7835.5 7835500 -0.8389 -0.8002 16 / 63 回归分析中，因变量Y和解释变量X的测量单位不同会造成回归结果的差异吗？ > GPDIB = 以2000年10亿(Billions)美元计国内私人总投资； GPDIM = 以2000 年百万(millions)美元计国内私人总投资； GDPB = 以2000年10亿(Billions)美元计GDP总值； GDPM = 以2000年百万 ( illi )美元计 总值 案例数据 案例数据 ( illi )美元计 总值 huhuaping@ 第5章：一元回归：模型形式扩展 5.2 尺度与测量单位 17 / 63 把某一测量单位下的回归模型，变换为另一测量单位的回归模型： Yi = ^ β1 + ^ β2Xi + ei Y ∗ i = ^ β ∗ 1 + ^ β ∗ 2Xi + e∗ i 尺度因子： 分别表示为Y和X的尺度因子。 ω1; ω2 Y ∗ i = ω1Yi X∗ i = ω2Xi 如果 都是以10亿(billion)美元计量的，我们把它们改为用百万(million)美 元去度量，就会有： (Yi; Xi) Y ∗ i = 1000Yi; X∗ i = 1000Xi; ω1 = ω2 = 1000 尺度变换 尺度变换 huhuaping@ 第5章：一元回归：模型形式扩展 5.2 尺度与测量单位 18 / 63 进行数据转换，新模型的OLS估计量如下： Yi = ^ β1 + ^ β2Xi + ei Y ∗ i = ^ β ∗ 1 + ^ β ∗ 2Xi + e∗ i Y ∗ i = ω1Yi; X∗ i = ω2Xi; e∗ i = ω1ei ^ β2∗= ⇐var(^ β ∗ 2) = ^ β ∗ 1 = ¯ ¯ ¯ ¯ Y ∗−^ β ∗ 2 ¯ ¯ ¯ ¯ ¯ X ∗ ⇐var(^ β ∗ 1) = ⋅σ∗2 ^ σ∗2 = ∑x∗ i y∗ i ∑x∗2 i σ∗2 ∑x∗2 i ∑X∗2 i n ∑x∗2 i ∑e∗2 i n −2 OLS估计 OLS估计 huhuaping@ 第5章：一元回归：模型形式扩展 5.2 尺度与测量单位 19 / 63 进行数据转换，两个模型下OLS估计量有如下关系： Yi = ^ β1 + ^ β2Xi + ei Y ∗ i = ^ β ∗ 1 + ^ β ∗ 2Xi + e∗ i Y ∗ i = ω1Yi; X∗ i = ω2Xi; e∗ i = ω1ei ^ β ∗ 1 = (ω1) ^ β1; ^ β ∗ 2 = ( ) ^ β2 var(^ β ∗ 2) = ( ) 2 var(^ β2); var(^ β ∗ 1) = ω2 1 var(^ β1) ^ σ∗2 = ω2 1^ σ2 r2 xy = r2 x∗y∗ ω1 ω2 ω1 ω2 OLS估计 OLS估计 huhuaping@ 第5章：一元回归：模型形式扩展 5.2 尺度与测量单位 20 / 63 Yi = ^ β1 + ^ β2Xi + ei Y ∗ i = ^ β ∗ 1 + ^ β ∗ 2Xi + e∗ i Y ∗ i = ω1Yi; X∗ i = ω2Xi; e∗ i = ω1ei 模型对比，得出如下主要结论： ，即尺度因子相等时，斜率系数及其标准误不受尺度从（ ）到（ ）的影响。截距及其标准误却放大或缩小至 倍。 ω1 = ω2 Yi, Xi Y ∗ i , X∗ i ω1 尺度不变 ， 尺度因子 变化，那么，斜率和截距系数以及它们各自的标 准误都要乘以同样的因子 。 Xi ω2 = 1 Yi ω1 ω1 尺度不变 ，而 尺度因子 变化，那么，斜率系数及其标准误都要乘以因 子 ，而截距系数及其标准误不变。 Yi ω1 = 1 Xi ω2 1/ω1 相关结论 相关结论 huhuaping@ 第5章：一元回归：模型形式扩展 5.2 尺度与测量单位 21 / 63 GPDI和GDP都以十亿美元计算： ˆ GPDIB = −926.09 + 0.25GDPB (t) (−7.9590) (19.5824) (se) (116.3577) (0.0129) (fitness) R2 = 0.9648; ¯ R2 = 0.9623 F ∗= 383.47;p = 0.0000 GPDI和GDP都以百万美元计算： ˆ GPDIM = −926090.39 + 0.25GDPM (t) (−7.9590) (19.5824) (se) (116357.6965)(0.0129) (fitness) R2 = 0.9648; ¯ R2 = 0.9623 F ∗= 383.47; p = 0.0000 GPDI以十亿美元，而GDP以百万美 元: ˆ GPDIB = −926.09 + 0.00GDPM (t) (−7.9590) (19.5824) (se) (116.3577) (0.0000) (fitness) R2 = 0.9648; ¯ R2 = 0.9623 F ∗= 383.47;p = 0.0000 GPDl以百万美元而GDP 以十亿美元： ˆ GPDIM = −926090.39 + 253.52GDPB (t) (−7.9590) (19.5824) (se) (116357.6965)(12.9465) (fitness) R2 = 0.9648; ¯ R2 = 0.9623 F ∗= 383.47; p = 0.0000 案例分析结果 案例分析结果 22 / 63 5.3 标准化变量回归 5.3 标准化变量回归 huhuaping@ 第5章：一元回归：模型形式扩展 5.3 标准化变量回归 23 / 63 假设如下双变量回归： Yi = ^ β1 + ^ β2Xi + ei 对Y和X作如下标准化变换，得到相应的标准化变量： Y ∗ i = ; X∗ i = Yi −¯ ¯ ¯ ¯ Y SY Xi −¯ ¯ ¯ ¯ ¯ X SX 标准化变量的特征是：其均值总是0 和标准差总是1。 标准化变量回归 标准化变量回归 huhuaping@ 第5章：一元回归：模型形式扩展 5.3 标准化变量回归 24 / 63 得到如下新的双变量回归模型: Y ∗ i = ^ β ∗ 1 + ^ β ∗ 2Xi + e∗ i = ^ β ∗ 2Xi + e∗ i 对标准化的回归子和回归元做回归，截距项总是零! 实际上变成了过原点回归模型! 标准化变量回归 标准化变量回归 huhuaping@ 第5章：一元回归：模型形式扩展 5.3 标准化变量回归 25 / 63 模型比较与结论： 第一，由于标准化回归本质上是一个过原点回归，而我们在已经指出通常过原点回归 的不能使用 ，所以我们就没有给出其 值。 r2 r2 第二，传统模型的系数与这里的系数之间存在一种有趣的关系。在双变量情形中，这 种关系如下（证明过程略：自学练习题！）： ^ β ∗ 2 = ^ β2 SX SY 第三，在多元回归中，变量标准化可以去除多个自变量之间数量尺度(量纲) 的差别，因而具有一定的优点！ 标准化变量回归 标准化变量回归 huhuaping@ 第5章：一元回归：模型形式扩展 5.3 标准化变量回归 Showing 1 to 8 of 16 entries Previous 1 2 Next Year GPDIB GDPB GPDIB_std GDPB_std 1990 886.6 7112.5 -1.2942 -1.3459 1991 829.1 7100.5 -1.4624 -1.3550 1992 878.3 7336.6 -1.3185 -1.1768 1993 953.5 7532.7 -1.0986 -1.0287 1994 1042.3 7835.5 -0.8389 -0.8002 1995 1109.6 8031.7 -0.6421 -0.6521 1996 1209.2 8328.9 -0.3508 -0.4277 1997 1320.6 8703.5 -0.0250 -0.1450 26 / 63 下面我们对以十亿美元计的GPDIB和GDPB进行标准化数据变换： 标准化数据变换 标准化数据变换 huhuaping@ 第5章：一元回归：模型形式扩展 5.3 标准化变量回归 27 / 63 GPDI和GDP都以十亿美元计算： ˆ GPDIB = −926.09 + 0.25GDPB (t) (−7.9590) (19.5824) (se) (116.3577) (0.0129) (fitness) R2 = 0.9648; ¯ R2 = 0.9623 F ∗= 383.47;p = 0.0000 标准化变量后的模型估计： ˆ GPDIBstd = + 0.98GDPBstd (t) (20.2697) (se) (0.0485) (fitness) R2 = 0.9648; ¯ R2 = 0.9624 F ∗= 410.86; p = 0.0000 OLS比较 OLS比较 28 / 63 回归模型的函数形式 回归模型的函数形式 对数线性模型 对数线性模型 半对数模型 半对数模型 倒数模型 倒数模型 29 / 63 5.4 对数线性模型 5.4 对数线性模型 huhuaping@ 第5章：一元回归：模型形式扩展 5.4 对数线性模型 30 / 63 指数回归模型（exponential regression model） Yi = β1Xβ2 i eui 可化为： ln Yi = ln β1 + β2 ln Xi + ui ln Yi = α + β2 ln Xi + ui ⇐α = ln β1 这种模型被称为对数-对数(log-log)，双对数(double-log)或对数一线性(log-linear) 模型。进而有： Y ∗ i = α + β2X∗ i + ui ⇐[Y ∗ i = ln Yi; X∗ i = ln Xi] 从而可用OLS方法可以得到BLUE估计量： Y ∗ i = ^ α + ^ β2X∗ i + ei 对数线性模型的形式 对数线性模型的形式 huhuaping@ 第5章：一元回归：模型形式扩展 5.4 对数线性模型 31 / 63 双数线性模型： ln Yi = α + β2 ln Xi + ui Y ∗ i = ^ α + ^ β2X∗ i + ei ⇐^ α = ln ^ β1 β2 = = = d(ln Y ) d(ln X) dY 1 Y dX 1 X dY /Y dX/X 斜率就是Y对X的弹性！如果Y代表商品需求量Q，X代表商品价格P，则就表示该 商品的需求价格弹性。 对数线性模型：学会如何测度弹性 对数线性模型：学会如何测度弹性 huhuaping@ 第5章：一元回归：模型形式扩展 5.4 对数线性模型 32 / 63 双数线性模型有如下性质： Y对X的弹性在整个研究范围内是常数，一直为 ，因此这种模型也称为不变弹性模 型(constant elasticity model)。 β2 虽然 和 是无偏估计量，但是进入原始模型的参数 的估计值 却是有偏估计， 而且 。 ^ α ^ β2 β1 ^ β1 β1 = antilog^ α 学会如何测度弹性 学会如何测度弹性 huhuaping@ 第5章：一元回归：模型形式扩展 5.4 对数线性模型 Showing 1 to 5 of 15 entries Previous 1 2 3 Next obs EXPDUR PCEXP ln_expdur ln_pcexp 2003-I 971.4 7184.9 6.8787 8.8797 2003-II 1009.8 7249.3 6.9175 8.8887 2003-III 1049.6 7352.9 6.9562 8.9029 2003-IV 1051.4 7394.3 6.9579 8.9085 2004-I 1067 7479.8 6.9726 8.9200 33 / 63 耐用品支出与个人消费总支出的关系： 其中：PCEXP=个人消费支出, EXPDUR=耐用品消费支出，单位10亿美元（按 2000年价格计） 耐用品消费案例 耐用品消费案例 huhuaping@ 第5章：一元回归：模型形式扩展 5.4 对数线性模型 34 / 63 假设我们想求出耐用品支出对个人消费总支出的斜率。 将耐用品支出相对于个人消费总支出做散点图： 耐用品消费案例 耐用品消费案例 huhuaping@ 第5章：一元回归：模型形式扩展 5.4 对数线性模型 35 / 63 假设我们想求出耐用品支出对个人消费总支出的弹性。 将耐用品支出的对数相对于个人消费总支出的对数做散点图： 耐用品消费案例 耐用品消费案例 huhuaping@ 第5章：一元回归：模型形式扩展 5.4 对数线性模型 36 / 63 耐用品消费案例中，我们可以实证得到如下的双对数模型： ˆ log(EXPDUR) = −7.54 + 1.63log(PCEXP) (t) (−10.5309) (20.3152) (se) (0.7161) (0.0801) (fitness) R2 = 0.9695; ¯ R2 = 0.9671 F ∗= 412.71;p = 0.0000 耐用品消费案例 耐用品消费案例 37 / 63 5.5 半对数模型 5.5 半对数模型 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 Showing 1 to 4 of 15 entries Previous 1 2 3 4 Next 劳务支出数据 obs t EXPSERVICES ln_expservices 2003-I 1 4143.3 8.3292 2003-II 2 4161.3 8.3336 2003-III 3 4190.7 8.3406 2003-IV 4 4220.2 8.3476 38 / 63 怎样测量增长率？经济学家、企业人员与政府常常对于求出某些经济变量的增长 率感兴趣，如人口、GNP、货币供给、就业、生产力、贸易赤字等。 Yt = Y0(1 + r)t ＝时期t的劳务实际支出； ＝劳务实际支出的初始值(为2002年第四季度末的 值)；r是Y的复合增长率。 Yt Y0 线性到对数模型 线性到对数模型 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 39 / 63 半对数模型(semilog models)： 线性到对数模型(log-lin model)：只有回归子Y取对数 对数到线性模型(lin-log model)：只有回归元X取对数 线性到对数模型 线性到对数模型 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 40 / 63 半对数模型的形式： Yt = Y0(1 + r)t ln Yt = ln Y0 + t ln(1 + r) ln Yt = β1 + β2t ⇐[β1 = ln Y0; β2 = ln(1 + r)] ln Yt = β1 + β2t + ut 斜率 的经济学含义： β2 β2 = = d ln Y dt dY /Y dt 恒定相对增长率模型：上述模型描述了因变量Y的恒定相对增长率 恒定相对增长模型： β2 > 0 恒定相对衰减模型： β2 < 0 线性到对数模型 线性到对数模型 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 41 / 63 散点图1 散点图1 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 42 / 63 散点图2 散点图2 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 43 / 63 半对数模型：线性到对数模型 ln Yt = β1 + β2t + ut ⇐[β1 = ln Y0; β2 = ln(1 + r)] ˆ log(EXPSERV ICES) = + 8.32 + 0.01t (t) (5186.2999) (39.9648) (se) (0.0016) (0.0002) (fitness) R2 = 0.9919; ¯ R2 = 0.9913 F ∗= 1597.18;p = 0.0000 ^ β2 = ln(1 + r) = 0.00705 r = antilog(^ β2) −1 = antilog(0.00705) −1 = 0.00708 表示瞬时增长率 ^ β2 = 0.00705 表示复合增长率 r = 0.00708 OLS估计 OLS估计 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 44 / 63 下面做一个对比模型。线性趋势模型：Y直接对时间t回归： Yt = β1 + β2t + ut ˆ EXPSERV ICES = + 4111.54 + 30.67t (t) (655.5628) (44.4671) (se) (6.2718) (0.6898) (fitness) R2 = 0.9935; ¯ R2 = 0.9930 F ∗= 1977.32;p = 0.0000 解释如下：在2003年第1季度至2006年第3季度期间，劳务支出以每季度约300 亿 美元的绝对速度(注意不是相对速度)增加，即劳务支出有上涨的趋势。 回归结果比较 回归结果比较 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 45 / 63 如果我们的目的是测量X的一个百分比变化时，Y的绝对变化量，则要用对数到 线性模型（lin-log model）。 Yt = β1 + β2 ln Xi + ui β2 = = = ΔY = β2 dY d ln X dY dX/X ΔY ΔX/X ΔX X 例如：恩格尔支出(Engel expenditure) 模型： “用于食物的总支出以算术级数增加，而总支出以几何级数增加。” 对数到线性模型(lin-log model) 对数到线性模型(lin-log model) huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 Showing 1 to 5 of 55 entries Previous 1 2 3 4 5 … 11 Next obs foodexp totalexp ln_totalexp 1 217 382 5.9454 2 196 388 5.9610 3 303 391 5.9687 4 270 415 6.0283 5 325 456 6.1225 46 / 63 食物支出（foodexp）与家庭总支出（totalexp）的关系： 家庭食物支出案例 家庭食物支出案例 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 47 / 63 原始数据作散点图： 家庭食物支出案例 家庭食物支出案例 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 48 / 63 对家庭总支出取对数ln(totalexp)，再做散点图： 家庭食物支出案例 家庭食物支出案例 huhuaping@ 第5章：一元回归：模型形式扩展 5.5 半对数模型 49 / 63 构建如下对数到线性模型： Yt = β1 + β2 ln Xi + ui 家庭食物支出案例的OLS估计结果如 下： ˆ foodexp = −1283.91 + 257.27log(totalexp) (t) (−4.3848) (5.6625) (se) (292.8105) (45.4341) (fitness) R2 = 0.3769; ¯ R2 = 0.3652 F ∗= 32.06; p = 0.0000 对比构建如下经典线性模型及其OLS 估计结果： Yt = β1 + β2Xi + ui ˆ foodexp = + 94.21 + 0.44totalexp (t) (1.8524) (5.5770) (se) (50.8563) (0.0783) (fitness) R2 = 0.3698; ¯ R2 = 0.3579 F ∗= 31.10; p = 0.0000 家庭食物支出案例 家庭食物支出案例 50 / 63 5.6 倒数模型 5.6 倒数模型 huhuaping@ 第5章：一元回归：模型形式扩展 5.6 倒数模型 51 / 63 形式： Yi = β1 + β2 ( ) + ui 1 Xi 特征：总有一条内在的渐近线！ a.平均固定成本(AFC)曲线 b.菲利普斯曲线（Phillips curve） c.恩格尔曲线（the Engel expenditure curve） X →∞; β2 ( ) →0; Y →β1 1 X 倒数模型 倒数模型 huhuaping@ 第5章：一元回归：模型形式扩展 5.6 倒数模型 Showing 1 to 5 of 64 entries Previous 1 2 3 4 5 … 13 Next obs CM PGNP rep_PGNP 1 128 1870 0.0005 2 204 130 0.0077 3 202 310 0.0032 4 197 570 0.0018 5 96 2050 0.0005 52 / 63 儿童死亡率（CM，千分数）与人均GNP（PGNP，1980年的人均GNP）的关系： 儿童死亡率案例 儿童死亡率案例 huhuaping@ 第5章：一元回归：模型形式扩展 5.6 倒数模型 53 / 63 原始数据作散点图： 儿童死亡率案例 儿童死亡率案例 huhuaping@ 第5章：一元回归：模型形式扩展 5.6 倒数模型 54 / 63 把PGNP取倒数 再作散点图： 1/PGNP 儿童死亡率案例 儿童死亡率案例 huhuaping@ 第5章：一元回归：模型形式扩展 5.6 倒数模型 55 / 63 构建如下倒数模型： Yi = β1 + β2 ( ) + ui 1 Xi 儿童死亡率案例倒数模型的OLS估计 结果如下： ˆ CM = + 81.79 + 27273.17repPGNP (t) (7.5511) (7.2535) (se) (10.8321) (3759.9992) (fitness)R2 = 0.4591; ¯ R2 = 0.4503 F ∗= 52.61; p = 0.0000 对比构建如下经典线性模型： Yt = β1 + β2Xi + ui 其OLS估计结果如下： ˆ CM = + 157.42 −0.01PGNP (t) (15.9893) (−3.5157) (se) (9.8456) (0.0032) (fitness)R2 = 0.1662; ¯ R2 = 0.1528 F ∗= 12.36; p = 0.0008 儿童死亡率案例 儿童死亡率案例 huhuaping@ 第5章：一元回归：模型形式扩展 5.6 倒数模型 Showing 1 to 5 of 47 entries Previous 1 2 3 4 5 … 10 Next year infrate unrate rep_unrate 1960 1.7182 5.5 0.1818 1961 1.0135 6.7 0.1493 1962 1.0033 5.5 0.1818 1963 1.3245 5.7 0.1754 1964 1.3072 5.2 0.1923 56 / 63 通货膨胀率（infrate，%）与失业率（unrate，%）的关系： 菲利普斯曲线 菲利普斯曲线 huhuaping@ 第5章：一元回归：模型形式扩展 5.6 倒数模型 57 / 63 原始数据作散点图： 菲利普斯曲线 菲利普斯曲线 huhuaping@ 第5章：一元回归：模型形式扩展 5.6 倒数模型 58 / 63 把失业率unrate取倒数 再作散点图： 1/unrate 菲利普斯曲线 菲利普斯曲线 huhuaping@ 第5章：一元回归：模型形式扩展 5.6 倒数模型 59 / 63 构建如下倒数模型： Yi = β1 + β2 ( ) + ui 1 Xi 菲利普斯曲线案例倒数模型的OLS估 计结果如下： ˆ infrate = + 7.37 −17.37repunrate (t) (4.1723) (−1.8212) (se) (1.7670) (9.5364) (fitness) R2 = 0.0686; ¯ R2 = 0.0479 F ∗= 3.32; p = 0.0752 对比构建如下经典线性模型及其OLS 估计结果： Yt = β1 + β2Xi + ui ˆ infrate = + 0.81 + 0.59unrate (t) (0.4642) (2.0377) (se) (1.7347) (0.2874) (fitness) R2 = 0.0845; ¯ R2 = 0.0641 F ∗= 4.15; p = 0.0475 菲利普斯曲线 菲利普斯曲线 60 / 63 5.7 函数形式的选择 5.7 函数形式的选择 1. 2. 3. 4. 5. 6. huhuaping@ 第5章：一元回归：模型形式扩展 5.7 函数形式的选择 61 / 63 选择适当模型时，需要一些技巧和经验： 模型背后的理论(如菲利普斯曲线〉可能给出了一个特定的函数形式。 最好能求出回归子相对回归元的变化率〈即斜率〉和回归子对回归元的弹性(见下页 ppt)。 所选模型的系数应该满足一定的先验预期。 有时多个模型都能相当不错地拟合一个给定的数据集。 通常不应该过分强调这个指标 在有些情形中，确定一个特定的函数形式不是那么容易，此时我们或许可以使用所谓 的博克斯-考克斯变换(Box-Cox transformations) 技巧和经验 技巧和经验 模型 方程 斜率 点弹性 平均弹性 models eq 线性模型 过原点模型 双对数模型 线性到对数模型 对数到线性模型 倒数模型 对数倒数模型 huhuaping@ 第5章：一元回归：模型形式扩展 5.7 函数形式的选择 dY dX ⋅ dY dX Xi Yi ⋅ dY dX ¯ X ¯ Y M1 Yi = β1 + β2Xi + ui β2 β2Xi/Yi β2 ¯ X/ ¯ Y M2 Yi = β2Xi + ui β2 β2Xi/Yi β2 ¯ X/ ¯ Y M3 ln(Yi) = β1 + β2ln(Xi) + ui β2Yi/Xi β2 β2 M4 ln(Yi) = β1 + β2Xi + ui β2Yi β2Xi β2 ¯ X M5 Yi = β1 + β2ln(Xi) + ui β2/Xi β2/Yi β2/ ¯ Y M6 Yi = β1 + β2/Xi + ui −β2/X2 i −β2/(XiYi) −β2/( ¯ X ¯ Y ) M7 ln(Yi) = β1 −β2/Xi + ui β2Yi/X2 i β2/Xi β2/ ¯ X 62 / 63 计算表一览 计算表一览 63 / 63 63 / 63 本章结束 本章结束
7448
https://amdc.impcas.ac.cn/web/masseval.html
The 2020 Atomic Mass Evaluation Atomic Mass Evaluation - AME 2020 The AME 2020 atomic mass evaluation has been published in Chinese Phys. C 45, 030002 (2021)and Chinese Phys. C 45, 030003 (2021). There are five main ASCII files associated with AME 2020: mass_1.mas20 - atomic masses [mass excess, binding energy/A, beta-decay energy, atomic mass] massround.mas20 - atomic masses "rounded" version [mass excess, binding energy/A, beta-decay energy, atomic mass] rct1.mas20 - reaction energies, table 1 [S(2n), S(2p), Q(a), Q(2B-), Q(ep), Q(B-n)] rct2_1.mas20 - reaction energies, table 2 [S(1n), S(1p), Q(4B-), Q(d,a), Q(p,a), Q(n,a)] covariance.mas20 - variances and covariances of primary nuclides [note this file is a .zip file] Any work that will use these files should make reference to the original papers listed above, and not to the electronic files. Archives Return
7449
https://www.youtube.com/watch?v=648RKXII7Xs
Using Trig Ratios To Find x. mrmaisonet 68300 subscribers 67 likes Description 6457 views Posted: 21 Jun 2023 Join us in this video as we demonstrate finding missing side lengths of right triangles using trigonometric ratios. 2 comments Transcript: all right in this problem we have to figure out what the length of side x is of this right triangle given the length of this side which is 19 units and the measure of this angle which is 24 degrees so let's figure out which one of these trig ratios we have to use in order to find the length of side x all right one thing that I would do as a beginner is I would just label what each side of this triangle is relative to this angle for example we know that the longest side of a right triangle is the hypotenuse so I'm going to label this side H for hypotenuse now out of the three sides one of the three sides will not form the angle in question for example this leg right here does not form this 24 degree angle this line segment does and this line segment does and the one that is not part of forming that angle is considered your opposite side so we would say that side x is opposite the 24 degree angle so I'm going to label this side o for opposite which leaves us with one thing remaining this side right here so this must be adjacent relative to our 24 degree angle now because we are dealing with an X and we're dealing with this 19 here that means we must use the opposite light and the adjacent length and the ratio that forms opposite and adjacent would be a tangent ratio so what we're going to do is we're going to write tangent of 24 degrees is equal to the length of the side that is opposite which in this case is X we don't know what that is yet over the adjacent length which is 19 in this case all right now whenever you have the denominator's length what we're going to do is we're going to automatically send it to the other side we're going to multiply 19 by the tangent of 24 degrees basically we just isolated the variable X because we're dividing by 19 we multiply by 19 on both sides so multiplying this by 19 cancels out 19 leaving us with just X so finding 19 times the tangent of 24 degrees is going to give us the length of x all right so let's input this into a calculator and see what the value of x is so we take 19 and we're going to multiply it by the tangent of 24 degrees and then we hit enter and we come up with a value that's 8.45 and some change now we have to round to the nearest tenth so 8.45 would be rounded to 8.5 so we would say that side x is approximately equal to 8 point five units okay same thing in this problem you just have to figure out what the length of X is given this angle right here which is 62 degrees and the length of this side which is 60 units all right so let's label what each side is relative to this angle right here so we know that the longest side is the hypotenuse so I'm going to label that H right away all right and then we have these two sides that form this right angle one of them is going to be your opposite side relative to this angle and one of them is going to be adjacent now remember the one that does not form the angle in question is always going to be your opposite which leaves us with the adjacent angle all right so they give us X right here and they give us 60 right here so we have to use adjacent in hypotenuse and the ratio that we would use to figure out PSI of X is going to be cosine because that is equal to the adjacent length divided by the hypotenuse so what we're going to do is we're going to write cosine of 62 degrees is equal to the adjacent side which is X over the length of our hypotenuse which is 60. all right now what we have to do is isolate our variable X because we're dividing by 60 we multiply by 60. and we have to balance our equation by multiplying 60 on the other side all right so let's go ahead and type in 60 times the cosine of 62 degrees and that's going to give us 28.1682 and some change so we would round this to 28.2 so we would say that X is approximately equal to 28.2 units [Music]
7450
https://nsidc.org/learn/parts-cryosphere/frozen-ground-permafrost/why-frozen-ground-matters
Home Learn Parts of the Cryosphere Frozen Ground & Permafrost Why Frozen Ground Matters Frozen Ground & Permafrost Overview Science Why it Matters Quick Facts In this section Frozen ground and wildlife Frozen ground and people Frozen ground and the environment Impacts of climate change on permafrost Frozen ground and wildlife Frozen ground and people Frozen ground and the environment Impacts of climate change on permafrost Why it Matters Since permafrost contains lots of organic matter like frozen plants and, occasionally, animals within, it holds an enormous amount of carbon. The Arctic contains nearly one-third of Earth's stored soil carbon in frozen ground and permafrost. If it thaws, microbial activity will lead to a release of heat-trapping greenhouse gases, carbon dioxide and methane, to the atmosphere. As of 2020, estimates suggest as much as 2.5 times more carbon is locked within permafrost than there is in the global atmosphere. Loss of permafrost may radically change the water and carbon cycle, significantly affecting climate change through feedback. Methane is a much more potent greenhouse gas than carbon dioxide, although it does not remain in Earth's atmosphere as long. Once the organic matter within these landscapes decomposes and releases these greenhouse gases into the atmosphere, there is no putting them back. Permafrost also lies beneath the Arctic Ocean, known as submarine or subsea permafrost, where vast reserves of geological methane are hibernating in an icy form known as methane clathrates. Scientists are concerned that, as the subsea permafrost thaws, this geologic methane could be quickly released to the atmosphere. Although permafrost can be thousands of years old, it is sometimes newly formed or about to thaw, and it often exists close to the melting point, making it particularly sensitive to even minor increases in temperature. In the case of subsea permafrost, changes in ocean temperature can lead to thawing. So despite its name, instability is a part of permafrost. In some areas, permafrost contains up to 80 percent ice. When this ice melts, the ground can collapse. This jeopardizes structures, triggers landslides, and feeds climate change feedbacks. Knowing where permafrost lies influences how buildings, roads, and other structures need to be built. For instance, the Trans-Alaska Pipeline System has 676 kilometers (420 miles) of above ground segments, where permafrost lies. As frozen ground thaws, pipelines, buildings, roads, airports, and factories are likely to be vulnerable to structural damage, requiring significant investments in repair and maintenance. Understanding permafrost, therefore, is important to civil engineering and architecture. It also determines the way of life for many communities from building cool cellars to store perishable food to how resources are delivered to those living in the Arctic and beyond. Carbon sink or source Plants consume carbon dioxide through photosynthesis. So, a greater supply of carbon dioxide has the potential to boost plant growth—particularly with extended growing seasons and shrinking permafrost in the Arctic. Some climate projection studies suggest that carbon dioxide released by thawing permafrost will be at least partially offset by increases in productivity of Arctic plants. But unlike model projections, a survey conducted in 2016 of 98 permafrost-region experts concluded that plants will not come to the rescue. Water stress will hamper plant growth, and wildfires on boreal forest may cause a four-fold increase in carbon release into the atmosphere. As a result, Earth's permafrost will likely become a carbon source rather than a carbon sink by 2100. Some recent studies suggest that this has already happened. Frozen ground and wildlife Animals living on frozen ground face special challenges: Food is limited, water is frozen, and the weather is extreme. Yet, animals thrive in these ecosystems. During winter, some animals migrate to places with easier living conditions, while others hibernate and sleep the season away in warm burrows. Some are so well adapted to the cold that they remain active throughout the winter. For example, the Arctic fox lives comfortably in areas with frozen ground. Like squirrels, the Arctic fox stores food such as bird eggs in the permafrost during summer. The eggs remain edible for up to a year. During the winter, the fox always has something to eat, even when the fox's favorite food, the collared lemming, is not available. The Arctic hare has also adapted. The Arctic hare does not build burrows underground, but rather builds its shelter in grass nests high up among rocks. Many insects, including mosquitoes, survive temperatures below freezing in regions of permafrost. With the arrival of spring, melted snow and rain pools on the surface of permafrost, forming bogs, marshes, and wetlands. Slow-moving and standing water are ideal breeding grounds for mosquitoes. The flat interior and south-central Alaska, for instance, is notorious for swarms of mosquitoes so thick they appear as dark clouds. Only females suck blood to nourish their eggs. A swarm of mosquitoes can siphon a quart of blood per week from hoofed animals like caribou, musk oxen, and reindeer. They can kill caribou calves and even adult caribou. Caribou are said to be always roaming, trying to escape the mosquitoes. Flies, deer flies, and tiny biting midges also breed in the marshy swamps above the permafrost. These insects are also important food for birds, but like mosquitoes, can also make life uncomfortable for animals and humans. After lying dormant through the winter, mosquito eggs hatch when waters become warm enough. Their wiggly larvae are a great food source for birds. However, as the Arctic continues to warm, mosquitoes hatch earlier and grow faster, avoiding the vulnerable state as bird prey. With lifecycles and migrations going out of sync, birds are missing the chance to feast on larvae. Therefore, climate change may boost mosquito populations in the Arctic. The caribou, related to deer and other hoofed animals, travel hundreds of miles to forage on the grasses and other foods found in the bogs and marshes. In the treeless tundra, caribou raise their calves safely, away from the predators that need cover to hunt. Predators can still hunt the caribou, but they are less successful than if they had places to hide and stalk the caribou. Brown bears, which include the famous and fearsome grizzly bear, are some of the Arctic's best-known large mammals. Brown bears fill up on food in the summer and fall to prepare for hibernation in the winter, sleeping for several months. This helps them conserve energy during the long winters, when food and water are scarce in permafrost and frozen ground habitats. If high latitude ecosystems change the timing of spring because of earlier and higher temperatures, brown bears may miss out on early spring food. Frozen ground and people People living in the Arctic have adjusted to life with permafrost. Generations in Alaska and Russia have stored their food in permafrost. Building such an underground refrigerator is no small task since the frozen ground is as hard as concrete. Thawing permafrost has tangible, often visible, and potentially dangerous effects. When the ground beneath turns to a soggy, muddy mess, buildings tilt, roads buckle and dip, and other infrastructures become unstable. Roads, bridges, railroads, pipelines, and other types of transportation infrastructure are vulnerable when permafrost thaws, needing constant repairs to keep them safe. There are several ways to stop road damage. Engineers sometimes replace soil under roads with gravel so that water drains better and there is less frost heave. Builders can elevate structures above the ground surface. Experts recommend painting roads white to reflect more heat and keep them cooler. A cooler road surface helps prevent frozen ground from thawing underneath. In some places with permafrost, the top layer of the ground thaws during the summer, but the water does not drain away. These wet areas are too marshy for roads. To solve this problem, people only drive in these areas during the winter. Engineers build ice roads on top of the frozen marshes. The ice on the roads is at least 100 centimeters (about 40 inches) thick. Trucks weighing up to 64 metric tons (70 tons) can drive (slowly!) across and haul supplies to mines and drill sites in northern Canada and Alaska. Ice roads are also built on frozen lakes for winter travel. In summer, the roads melt, so they must be rebuilt each winter. For people living in the Tibetan Plateau in southwestern China, the high altitude makes travel difficult between cities in Tibet and other major cities in China. The only way to get to some parts of the Tibetan Plateau was once to fly. In 2006, to link Lhasa, the capital city of Tibet, with China, the Chinese government opened the Qinghai-Tibet Railway, the world’s highest train route, reaching altitudes over 5,000 meters (16,400 feet). The railway line, which cost US$4.2 billion to build, required designers and engineers to tackle three main challenges during construction: the fragile ecosystem, the lack of oxygen, and permafrost. The railroad stretches 1,100 kilometers (684 miles), nearly the distance between New York City and Chicago. Of that distance, 960 kilometers (597 miles) sits at 4,000 meters (13,100 feet). Special construction methods had to be used to build on the permafrost, as running a railroad could create heat that would thaw the permafrost beneath, buckling its tracks. To make sure that the ground stayed frozen, Chinese engineers used crushed rock to insulate the ground and built high bridges to keep the train tracks above the permafrost. When ice forms, it expands. So, constructing infrastructure on frozen ground is difficult. For instance, frost heaves lift the ground and everything on top of it. Also, the heating systems within buildings can potentially thaw the ground beneath the building. This would cause the ground to sink. Engineers sometimes solve this problem by preventing the ground under the building from getting warm. They put the building on top of a steel frame, a few feet above the ground, so cold air can flow under the house. The cold air stops the permafrost from thawing. Another way to stop damage from thawing permafrost is to thaw the ground first. This method makes the ground more stable to build on. Then there is no danger of the ground beneath the new structure refreezing because the structure keeps the ground from freezing. Frozen ground and drinking water In areas with seasonally frozen ground, the groundwater under the surface usually does not freeze, making drinking water accessible. Plus, lakes, rivers, and reservoirs can provide water. Permafrost, however, is more challenging as most of the groundwater is frozen. Typically, any water that is liquid may not taste good. The ice in the soil pushes its minerals out. These minerals get concentrated in any liquid water in the soil, affecting the taste of the water. In places of discontinuous permafrost, people do find water. They drill through areas of unfrozen ground to get to the groundwater. In places with large, continuous stretches of permafrost, finding water takes a lot of effort. People can sometimes get water from nearby lakes and rivers or by melting ice or snow, but they cannot get liquid water directly out of the ground in the winter. Some towns build water pipes from the water supply to the buildings. However, the pipes have to be protected so that water inside the pipes does not freeze, and so that the ground around the pipe does not thaw. Sometimes, engineers avoid this problem by building pillars to keep the protected pipes above ground. Permafrost effect on oil and natural gas Alaska, Canada, and Russia all have important deposits of oil and natural gas that give our cities, towns, and farms energy. Getting oil and natural gas out of the ground and to the people who use it is a huge effort. Drilling deep wells for oil and gas can thaw permafrost. If the permafrost thaws, the wells can collapse. Drilling companies put their equipment on special concrete pads built to prevent the ground underneath from thawing. Concrete well liners prevent wells from collapsing. Companies also use special drilling liquids that do not freeze as quickly as water does to lubricate the drill bits. Once the oil or gas is out of the ground, companies need to move it to where it will be used. In 1977, oil companies built the Alaska Pipeline. The pipeline takes oil 1,299 kilometers (800 miles) through Alaska to the shipping port of Valdez. Building the Alaska Pipeline across permafrost required special consideration. Oil in the pipeline must be kept above 60°C (140°F) so that it flows easily. The oil would be warm enough to thaw the permafrost and cause the pipeline to sink and break, so engineers built the pipeline above the ground in areas of permafrost. The pipeline can run underground in sections that go through seasonally frozen ground. They also constructed devices to put the extra heat into the air rather than letting it go into the ground. Frozen ground and the environment Plants can grow in extreme conditions, including frozen ground. Plants need sunlight, nutrients from the soil, and water to live. In some places in the Arctic, the ground is frozen most of the year, and months go by without any sunlight. So, how do these plants survive? Plants living in these areas must survive the winter with little water. They push their roots deep down under the frozen topsoil to find liquid groundwater. Plants that grow on seasonally frozen ground can grow quite tall because under the layer of frozen ground, the deep soil layer can support them. When snow melts and seasonally frozen ground thaws in the spring and summer, bogs can form. Trees thrive in the rich, moist soil. In the Arctic, huge forests cover areas of seasonally frozen ground. The forests are called boreal forests. Sometimes they are called taiga, the Russian word for swampy, moist forest. Typically, trees in the boreal forest are conifers. Conifers have small needle-like leaves that may have a waxy coating. Confers save water because their leaves do not have much surface area. For example, an oak leaf is broad, thin, and delicate, with a large surface area, causing much more water loss from transpiration than thin pine needles. By conserving water, conifers do not need to get as much water out of the ground. Surviving in permafrost, rather than frozen ground, is a much larger challenge for plants. On top of extremely low air temperatures and frozen water, permafrost contains only a thin layer of topsoil. Tundra is a type of biome where low temperatures, short growing seasons, and larger stretches of permafrost hinder tree growth. The word tundra is a Finnish word referring to a treeless plain. Tundra is found at high latitudes and at high altitudes, where the permafrost has a very thin active layer that cannot support a tree's roots. Tundra areas are some of the coldest and driest on Earth. Strangely, even though the tundra does not get much rain or snow, it often has many shallow lakes and ponds, wetlands, and swamps during warm seasons. This is because permafrost acts like a barrier, preventing water from draining through it. The grasses, lichens, and shrubs that do grow on the tundra make a thick and colorful carpet. Not all permafrost is tundra. Boreal forests do sometimes move into areas where areas of permafrost are mixed with seasonally frozen ground. The plants of the tundra and the permafrost underneath are in balance. Plants growing on the surface absorb solar energy, protecting the permafrost and preventing it from thawing. The permafrost keeps melted water near the surface, where plants need it. However, this balance is fragile. If the plant cover is damaged, the permafrost can thaw. Thawing permafrost can make the ground collapse and disturb and deepen the active layer. This can also affect the plants and animals that have lived there for many years. If the permafrost is damaged, plants also suffer. Fires, floods, or unusually high temperatures can thaw it. Trees growing in boreal forests are threatened. Sometimes, their roots become so weak that trees tilt and fall over, known as a drunken forest because of how trees topple onto one another. Impacts of climate change on permafrost As Earth's climate warms, permafrost and frozen ground thaw, shrink in their extent, and/or disappear entirely. In the Northern Hemisphere, the amount and thickness of seasonally frozen ground is decreasing. Above permafrost, the active layer, where the ground freezes and thaws each year, is getting thicker, meaning the permanently frozen ground layer is thinning. In Russia and China, scientists have found that the active layer has become much thicker since the early 1980s: In Siberia, it is up to 25 centimeters (10 inches) thicker. On the Tibetan Plateau, it is up to 100 centimeters (40 inches) thicker. The Arctic Report Card: Update for 2017 reported continued increases in active layer thickness since the 1990s, at sites measured in North America, Eurasia and Greenland. Likewise, more pockets of unfrozen ground known as taliks are forming within frozen ground. Eventually, very little permafrost may be left, depending on the degree of warming. Likewise, seasonally frozen ground and permafrost areas will not reach as far south as they do now in the Northern Hemisphere. As the active layer becomes thicker, the landscape may change. The ground can become weak. In hilly areas, thawing can cause landslides. In some areas, thawing changes the landscape, creating slumping ground, unstable forests, and shallow lakes. These landforms are examples of thermokarst features. Scientists have found that there is now 10 percent less frozen ground in the Northern Hemisphere than in the early twentieth century. Ten percent equals more than 5 million square kilometers (2 million square miles), about two-thirds the size of the continental United States. As of 2021, Earth has warmed about 1°C (1.8°F) since preindustrial levels. A 2017 study suggests that with every additional 1°C (1.8°F) warming, another 4 million square kilometers (1.5 million square miles) of permafrost will disappear. This is about 20 percent higher than previous studies, according to the scientists. The study also states that under a business-as-usual greenhouse-gas-emission scenario, which would result in a 6°C (10.8°F) warming sometime after 2100, nearly all of Earth’s permafrost would likely disappear. All of these changes will affect how plants grow, and how carbon and water will cycle through the environment. Without permafrost, water will drain away or evaporate into the air. Wetlands will dry up. Fewer plants will grow, and with reduced resources, fewer animals will be able to survive. People may also find it harder to find water and food in regions where permafrost has disappeared. Moreover, the increased abundance of greenhouse gases into the atmosphere will only exacerbate warming and continue permafrost thaw. Warming leads to cascading impacts on arctic land animals. Swedish Saami, a group known for reindeer herding, have observed that since the mid-1990s thawing frozen ground in summer grazing areas, leads to soggy falls and rotten spring vegetation. Permafrost thaw, erosion, and community collapse About 500 people live in Shishmaref, a small town on a barrier island near the Bering Strait, just below the Arctic Circle. The island is only about a quarter of a mile wide, and it is shrinking. The Iñupiaq people lived nomadic lives in the area for thousands of years until the government forced them into stationary lives. Then, they witnessed as their town started to literally fall into the sea. Shishmaref lies on a permafrost island along the coast of the Bering Sea. In the past, the ground stayed frozen, and sea ice hugged the shores. Sea ice protected the shore from waves even in the summer. However, since the 1980s, summer sea ice has become much more unpredictable, with the summers of 2018 and 2019 setting record lows for the region. Now, waves batter the shoreline in summer, washing away the coastline. Every year, about 7 meters (23 feet) of Shishmaref washes into the sea. The people of Shishmaref have moved their buildings away from the shore, put rocks and sandbags down to protect the coast, and built seawalls. But the waves are still washing the land away. Villagers have voted three times to relocate: in 2016, 2002, and as far back as 1973. Though in 1973, climate change was not yet at the forefront, worries about erosion were already evident, only to accelerate a few decades later. Though Shishmaref and other towns have been engaged in plans to move, moving an entire community is complicated and costly. According to one report put together by the US Army Corps of Engineers, the cost of moving Shishmaref would be somewhere around $179 million. The deep thaw: what’s locked within? Mammoth bones are surfacing in the Russian Far East—so many that people have begun selling the tusks as a substitute for elephant ivory. Permafrost thaw has far-reaching consequences for all life on the planet, including humans. Just like thawing meat out of your freezer begins to rot after a few days, so does the organic matter trapped within frozen soil. Rising soil temperatures activate the microorganisms once frozen within. Once turned on, they degrade the plant and animal matter within the soil, releasing loads of carbon and methane and intensifying global warming. Right now, Earth's atmosphere contains about 850 gigatons of carbon. A gigaton is one billion tons—about the weight of one hundred thousand school buses. Scientists estimate that there are about 1,400 gigatons of carbon frozen in permafrost. That means that permafrost holds 2.5 times the carbon that is currently in the atmosphere. Once released, a cycle ensues of warming soil, increasing heat-trapping gasses into the atmosphere, and raising global temperatures. Scientists are still studying the amount of carbon stored in permafrost and how quickly it might break down. For example, scientists found a type of permafrost that is rich in matter from dead plants and animals, called yedoma. Yedoma is a type of relict permafrost, left over from the Pleistocene epoch. It mostly exists in Siberia in northern Russia. If all yedoma were to thaw, it could release a very large amount of carbon to the atmosphere. Harmful once-frozen bacteria wake with the thaw. In 2016, in remote Siberia on the Yamal Peninsula, a 12-year-old boy died of an unknown disease. One hundred other people were hospitalized, 20 were diagnosed with the same disease, and more than 2,300 reindeer were found dead. According to Russian officials, thawed permafrost was the cause. Previously immobile spores of Bacillus anthracis seeped out of the thawing ground and into nearby waters and soil, and down into the food supply. Anthrax infection typically attacks livestock and wild animals. Usually, anthrax bacteria enter the body through a wound in the skin. A person can also become infected by eating contaminated meat or inhaling the spores. Inhaled anthrax is more difficult to treat and can be fatal. Researchers have predicted that higher temperatures, combined with permafrost thaw, could trigger the release of dormant diseases. As warming increases, the spread of thawing permafrost likewise increases. The release of dormant diseases depends on several factors: Where the pathogens are located in terms of permafrost depth and global location How high temperatures rise And the strength of pathogen variants For instance, not all pathogens can survive extremely cold temperatures. Anthrax, however, can. It is possible that viruses once thought to have been eradicated by humans could return on a large scale, like smallpox or the Spanish flu. A 2014 study showed how two scientists resurrected a virus from a 30,000-year-old piece of ice in Siberia. Although this very large virus targets amoebas, it was still infectious after all that time. The two scientists attempted to revive this virus after learning about Russian scientists who successfully grew an ancient plant from 30,000-year-old seeds. Ground squirrels buried a hoard of fruits and seeds, and the permafrost encapsulated them. In 2012, when the plant was resurrected, it was the oldest living plant ever revived. Mercury rising: thawing permafrost Researchers have discovered that thawing permafrost in the Northern Hemisphere traps another by-product of decaying matter. Mercury naturally occurs in the Earth’s crust and typically enters the atmosphere through volcanic eruptions. The element cycles between the atmosphere and ocean quickly. However, mercury deposited on land from the atmosphere binds with organic matter in plants. After the plants die, soil microbes eat the dead organic matter, releasing the mercury back into the atmosphere or water. In permafrost regions, however, the organic matter gets buried by sediment before it decays and becomes frozen into permafrost. Once frozen, the decay of organic matter stops, and the mercury remains trapped for thousands of years unless liberated by permafrost thaw. As long as the permafrost remains frozen, the mercury will stay trapped in the soil. However, with global temperatures rising, this seems unlikely. In a 2018 study, scientists measured mercury concentrations in cores of permafrost from Alaska. They used the data to estimate how much mercury has been trapped in Northern Hemisphere permafrost since the end of the last Ice Age about 12,000 years ago. Apparently, Northern Hemisphere permafrost regions contain 1,656 gigagrams of mercury (32 million gallons), or enough to fill 50 Olympic-sized swimming pools. These regions are the largest known reservoir of mercury on the planet. This amount is nearly twice as much mercury as all soils outside the northern permafrost region, the ocean, and the atmosphere combined. The researchers also found that of the 1,656 gigagrams of mercury, 863 gigagrams lie in the surface layer of soil that freezes and thaws each year (27 Olympic-sized swimming pools), and 793 gigagrams are frozen in permafrost (23 Olympic-sized swimming pools). The finding has significant implications for human health and ecosystems worldwide. For instance, once mercury is released because it is concentrated in these areas, it would pollute rivers, groundwater, and all life that depends on this water. In 2020, a NSIDC researcher published a study stating that the concentration of mercury in fish in Alaska’s Yukon River may exceed the Environmental Protection Agency’s mercury criterion by 2050. This first-of-its-kind research estimates potential releases of mercury from thawing permafrost in high and low greenhouse gas emissions scenarios. The researchers predict that by 2200, the mercury emitted into the atmosphere annually by thawing permafrost could compare with current global anthropogenic emissions under a high emissions scenario. The results indicate minimal impacts to mercury concentrations in water and fish for the low emissions scenario and large increases for the high emissions scenario. The high emissions scenario shows mercury releases to the atmosphere comparable to current anthropogenic emissions, with large increases in mercury concentrations in fish and water in the Yukon River. The low emissions scenario shows minimal releases to the atmosphere and small changes to mercury concentrations in fish and water. For the high emissions scenario, mercury concentrations could double in the Yukon River by 2100. For the low emissions scenario, mercury concentrations would likely increase by only about 14 percent and would not exceed EPA criterion by 2300. Methane: the other greenhouse gas Methane is a gas made up of one carbon atom and four hydrogen atoms. As organic matter decays, microbes digest it and produce either carbon dioxide or methane as waste. If there is oxygen available, the microbes make carbon dioxide. But if there is no oxygen available, they make methane. Most of the places where methane would form are swamps and wetlands. And many miles of wetlands are in the Arctic. When walking around in the Arctic tundra in the spring or summer, it is like sloshing through a giant sponge. When permafrost carbon turns into methane, it bubbles up through soil and water. On the way, other microorganisms eat some of it. But some methane makes it to the surface and escapes into the air. Like carbon dioxide, methane is a greenhouse house with heat-trapping properties. Methane lingers in the air about a decade on average, which is much less time than carbon dioxide, but it traps much more energy too. According to the US Environmental Protection Agency, methane is between 28 to 36 times more efficient at trapping heat than carbon dioxide over a 100-year period—the time period used for comparing different gases and their global warming potential. To put it simply, methane is much more powerful than carbon dioxide at warming the climate but breaks down faster. Human activities are responsible for about 60 percent of methane emissions. Methane seeps out during the production, processing, storage, transport, and use of coal, natural gas, and oil. It is the primary component of natural gas, used to heat many homes. Methane also enters the atmosphere through agriculture (including fermentation, manure management, and rice production), landfills, and wastewater treatment. There are two potential sources of methane in the Arctic. The first is called methyl clathrate. Methyl clathrates are methane molecules that are frozen into ice crystals. They can form deep in the earth or underwater, but it takes very special conditions, with high-pressure and low temperature, to make them. If the temperature or pressure changes, the ice that locks in the methane will break apart, and the methane will escape. Scientists are not sure how much methane is trapped in methyl clathrates, or how much is in danger of escaping. Estimating methane release from clathrates under the Arctic’s subsea permafrost is an active area of research. The other major source comes from thawing dead plants and animals that have been trapped in the frozen permafrost for thousands of years. As long as this organic matter remains frozen, methane will not escape. However, when permafrost thaws, the organic matter decays, releasing carbon dioxide and/or methane into the atmosphere. Thawing permafrost and sea level rise If all the permafrost in the world thawed, it could release enough water to raise global sea levels by 3 to 10 centimeters (1 to 4 inches). This might not seem like very much water, but it is enough that many cities along coastlines would need to build walls to keep the sea out. Both model projections and observations have pointed to current and future permafrost degradation. A study published in 2005 suggests that by 2100, as little as 1.0 million square kilometers (386,000 square miles) of near-surface permafrost might remain, increasing the freshwater discharge from rivers into the Arctic Ocean by 28 percent. This projection applied only to the top few meters of permafrost and depended upon climate and snow cover models. A 2022 study examining climate model output corroborated the general projection of the 2005 study: widespread permafrost loss by 2100. Field observations indicate that permafrost warmed up to 6°C (10.8°F) during the twentieth century. In January and April 2006, air temperatures on Svalbard reached more than 12°C (21.6°F) above the 1961 to 1990 average. On June 30, 2021, the temperature at Fort Smith, Northwest Territories, Canada, reached 39.9°C (103.8°F), a heatwave that could have broken records at much lower latitudes. Observations in Alaska found permafrost warming at most sites north of the Brooks Range from the Chukchi Sea to the border with Canada, coincident with statewide air-temperature warming beginning in 1976. The warming occurred primarily in the winter, with little summertime change. Close menu
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UKOxfordCambridgeLBSLSEImperialUCLBayesKing'sWarwickManchester Alliance Canada Top Master in Finance Colleges in CanadaYork SchulichQueen's SmithUBC SauderMcGillWaterlooHEC MontrealConcordiaSobeySimon Fraser Germany Top MFin Colleges in GermanyTUMFSFMWHU OttoHHL LeipzigEBS Australia Best MFin Colleges in AustraliaUniversity of MelbourneANURMIT Europe Top Masters in Finance Colleges in EuropeHEC ParisESSECEDHECIÉSEGESCPEMLYONGrenobleIEESADERotterdam Masters in Engineering Management USA Best colleges for Engineering Management in USAStanfordMITColumbiaCornellPurdueMichigan DearbornUSC ViterbiNorthwestern KelloggDartmouth TuckJohn HopkinsDuke FuquaTuftsNortheasternPenn StateUC Irvine Please screenshot this page and send it to info@mim-essay.com close Table of Contents Understanding GMAT Fractions GMAT Fractions to Memorize Common Fraction for GMAT Overview of Frequently Used Fractions Fractions in Word Problems Fractions in Ratio and Proportion Problems Converting Fractions to Decimals Simplifying Fractions for GMAT GMAT Fractions in Data Sufficiency Questions Common Mistakes Students Make with GMAT Fractions Practice Questions for GMAT Fractions GMAT Fractions in Word Problems GMAT Fractions in Ratio and Proportion Problems Comparing Fractions on the GMAT GMAT Fractions: Essential Tips, Practice Problems & Strategies A Complete Guide to Understanding and Solving GMAT Fractions Home Blogs Study Abroad GMAT Fractions: Essential Tips, Practice Problems & Strategies Last updated at 24 September 2024 10 minutes Study Abroad Table of Contents Understanding GMAT Fractions GMAT Fractions to Memorize Common Fraction for GMAT Overview of Frequently Used Fractions Fractions in Word Problems Fractions in Ratio and Proportion Problems Converting Fractions to Decimals Simplifying Fractions for GMAT GMAT Fractions in Data Sufficiency Questions Common Mistakes Students Make with GMAT Fractions Practice Questions for GMAT Fractions GMAT Fractions in Word Problems GMAT Fractions in Ratio and Proportion Problems Comparing Fractions on the GMAT Key Takeaways Key Fractions to Memorize: Familiarize yourself with commonly used fractions and their decimal equivalents to save time during the exam. Simplifying Fractions: Learn techniques to simplify fractions efficiently for faster problem-solving. Converting Fractions to Decimals: Understanding how to convert between fractions and decimals is critical for solving various GMAT questions. Comparing Fractions: Discover methods to compare fractions quickly without needing complex calculations. Fractions play a significant role in the GMAT Quantitative section, appearing in everything from problem-solving to data sufficiency questions. Understanding how to manipulate and simplify fractions quickly is essential for boosting your score. Whether it's converting fractions to decimals, comparing fractions, or solving complex word problems, mastering these concepts will give you a competitive edge on test day. This guide covers the key fraction-related skills you need to excel in the GMAT. Understanding GMAT Fractions Fractions are an essential part of the GMAT quantitative section. Whether you're solving word problems, comparing ratios, or simplifying complex equations, a strong grasp of fractions can make a significant difference in your performance. On the GMAT, you will often see questions requiring quick calculations involving fractions, so it's crucial to understand how they work and the common fractions you need to memorize. GMAT Fractions to Memorize Memorizing key fractions and their decimal equivalents can significantly improve your speed on the GMAT. While it’s possible to calculate these values during the test, knowing them by heart will allow you to focus on solving the problem instead of doing basic conversions. Here’s a deeper look at why memorization is essential and the key fractions to know. Why Memorizing Fractions Helps Many GMAT questions, especially in the Quantitative section, involve operations with fractions. Whether you’re dealing with ratio questions, percentage problems, or even algebraic expressions, having these fractions already in mind will make solving questions much faster. For example: Ratio problems: If a problem involves ratios, fractions like 1/3, 1/4, and 3/5 frequently appear. Knowing their decimal equivalents can help in calculating the final answer. Percentage problems: Percentages can often be expressed as fractions. For example, 25% is equal to 1/4, and 66.67% is 2/3. Instead of converting these during the test, knowing them beforehand will give you an edge. Knowing these conversions also reduces the risk of calculation errors, which are common when you’re under time pressure. Memorizing even a few key fractions can lead to quicker problem-solving, which is vital when you only have a limited time to answer each question. Key Fractions to Memorize Here is an extended list of fractions and their decimal equivalents that you should memorize for the GMAT. These are among the most commonly encountered in GMAT questions: | Fraction | Decimal Equivalent | --- | | 1/2 | 0.5 | | 1/3 | 0.333... | | 1/4 | 0.25 | | 1/5 | 0.2 | | 1/6 | 0.166... | | 1/8 | 0.125 | | 1/10 | 0.1 | | 2/3 | 0.666... | | 3/4 | 0.75 | | 3/8 | 0.375 | | 5/6 | 0.833... | | 7/8 | 0.875 | Please refer GMAT Quantitative: Fractions and Percents for detailed analysis of GMAT Fractions Common Fraction for GMAT In the GMAT Quantitative section, fractions are frequently used across different problem types like ratios, algebra, and word problems. Understanding how to manipulate common fractions is essential for answering these questions efficiently. Overview of Frequently Used Fractions Some fractions appear more regularly in GMAT problems than others. These fractions, often paired with percentages or ratios, are used in both problem-solving and data sufficiency questions. Below is a list of common fractions that every GMAT test-taker should be comfortable with: | Fraction | Decimal Equivalent | --- | | 1/2 | 0.5 | | 1/3 | 0.333... | | 2/3 | 0.666... | | 1/4 | 0.25 | | 3/4 | 0.75 | | 1/8 | 0.125 | | 7/8 | 0.875 | | 1/5 | 0.2 | | 3/5 | 0.6 | | 4/5 | 0.8 | Fractions in Word Problems Word problems often involve fractions in the context of dividing quantities or calculating parts of a total. For instance: “A company’s annual budget is $1,200,000. If 1/3 of the budget is allocated to marketing, how much is spent on marketing?” To solve this problem, you multiply $1,200,000 by 1/3. Knowing that 1/3 is 0.3333 makes this calculation much faster. The answer is $400,000. Similarly, percentage problems frequently involve fractions. For example, if a question asks what 25% of 320 is, recognizing that 25% is equivalent to 1/4 will help you solve the problem in seconds (1/4 of 320 = 80). Fractions in Ratio and Proportion Problems In ratio problems, you often need to convert ratios into fractions. For example, a question may ask for the total number of parts in a ratio of 3:2. This is equivalent to 3/5 and 2/5 of the total. Understanding how to work with these fractions allows for quicker and more accurate problem-solving. Here’s an example: “A recipe calls for 3 parts water to 2 parts flour. If you have 500 grams of flour, how much water should you use?” To solve this, you recognize that flour makes up 2/5 of the total mixture, so 500 grams represents 2/5 of the total weight. Solving for the total weight and subtracting the weight of the flour gives you the amount of water needed. By mastering the manipulation of these common fractions, you can solve these types of GMAT questions more efficiently and confidently. Converting Fractions to Decimals Being able to quickly convert fractions to decimals is a key skill for the GMAT. Not only does it help in solving fractions-based problems, but it’s also useful when dealing with percentage questions. Mastering fraction-to-decimal conversions can make many questions much faster to solve, especially when dealing with GMAT’s multiple-choice format. Quick Conversion Tricks There are some shortcuts and tricks you can use to easily convert common fractions to decimals. Here's a breakdown of some common conversions: | Fraction | Decimal Equivalent | Conversion Trick | --- | 1/2 | 0.5 | Divide 1 by 2 | | 1/3 | 0.333... | Divide 1 by 3, repeating decimal | | 1/4 | 0.25 | Divide 1 by 4 | | 1/5 | 0.2 | Divide 1 by 5 | | 1/8 | 0.125 | Divide 1 by 8 | | 3/4 | 0.75 | Multiply 0.25 by 3 | | 2/5 | 0.4 | Multiply 0.2 by 2 | Use of Conversion in GMAT Problem-Solving Fraction-to-decimal conversions are often useful in questions where you need to compare two values. For example, if you’re comparing 3/8 to 0.4, converting both into decimals will make it easier. In this case, 3/8 converts to 0.375, which is smaller than 0.4. Here’s an example: “Which is greater, 5/6 or 0.85?” By converting 5/6 to a decimal (which equals 0.8333), it’s clear that 0.85 is larger. These kinds of comparisons are frequent in GMAT problems, particularly in data sufficiency questions. Practice Conversions Below are a few practice problems to help you get comfortable with fraction-to-decimal conversions. Try solving these without using a calculator to speed up your mental math skills: Convert 7/8 to a decimal Convert 3/5 to a decimal Compare 2/3 and 0.66 – which is larger? Is 0.25 greater or smaller than 1/4? By practicing these conversions regularly, you’ll be able to answer GMAT questions involving fractions and decimals more confidently and quickly. Please refer GMAT Practice Questions with Fractions and Decimals for detailed analysis of GMAT Fractions Simplifying Fractions for GMAT Another key skill tested on the GMAT is the ability to simplify fractions. This often involves reducing fractions to their simplest form, making it easier to work with them in calculations. Simplifying fractions is especially useful in GMAT problem-solving and data-sufficiency questions, where clear and efficient calculations are necessary. The Importance of Simplifying Fractions On the GMAT, questions often involve complex calculations with fractions. Simplifying fractions early in the process makes these problems more manageable. For example, rather than multiplying large numbers, simplify the fractions first to reduce the numbers you're working with. For instance: “Solve for x: (6/8) (4/6) = x” Rather than multiplying 6 by 4 and 8 by 6, first simplify 6/8 to 3/4, and 4/6 to 2/3. Now, the problem becomes: “(3/4) (2/3) = x” The answer is 1/2, much easier to calculate when the fractions are simplified first. Techniques to Simplify Fractions The main technique for simplifying fractions is to divide both the numerator and the denominator by their greatest common divisor (GCD). Here’s how you can simplify fractions: Identify the greatest common divisor of the numerator and denominator. Divide both the numerator and denominator by the GCD. The resulting fraction is the simplified version. For example, to simplify 18/24: The GCD of 18 and 24 is 6. Divide both by 6: (18 ÷ 6) / (24 ÷ 6) = 3/4. We can help you get into your Dream Schools If you book a call, we will cover the following: Detailed Analysis of Profile Strengths and Weaknesses ⁠Ideal 8 Schools to target What GMAT Score do you really need ⁠Things you can do now to Improve your Profile How to improve your Admit and Scholarship chance Name Enter your Name Email ID Enter your Email Contact Number +91 Enter your Number Book a Free Session Now Your Appointment is being confirmed! (Step 1 of 2) Just give us one last detail and you are good to go! Which intake are you applying to? (This would be the year you will be starting your studies in) Please Choose Atleast One Option 2026 2027 and Later Submit Examples of Simplification in GMAT Problems Here are a few examples where simplifying fractions makes solving GMAT problems easier: Simplify 9/12 before multiplying it with another fraction (3/4). In a data sufficiency question, you may need to simplify ratios (e.g., 12:8 simplifies to 3:2). Simplify 15/35 in a word problem to 3/7 for easier calculations. Mastering this technique will make fraction problems more manageable and help improve your speed on the GMAT. GMAT Fractions in Data Sufficiency Questions The GMAT’s data sufficiency section is unique because it tests your ability to determine whether you have enough information to solve a problem, rather than actually solving it. Fractions often appear in data sufficiency questions, and being able to work with them efficiently is crucial for answering these questions correctly. How Fractions are Tested in Data Sufficiency In data sufficiency questions, fractions may appear in the form of ratios, parts of a whole, or percentages. The challenge is to determine whether the given statements provide enough information to solve the problem. For example, a typical data sufficiency question involving fractions might look like this: “If 3/4 of a tank is filled with water, and Statement (1) says that the tank’s total capacity is 100 liters, can you determine how much water is in the tank?” Here, knowing how to work with the fraction 3/4 will help you quickly determine that 3/4 of 100 liters is 75 liters, meaning the problem can be solved with the information provided in Statement (1). Approach to Solving Fraction-Related Data Sufficiency Questions To solve fraction-related data sufficiency questions, follow these steps: Step 1: Analyze the problem – Determine whether the fraction given is essential to the problem. Step 2: Check each statement independently – Assess whether each statement provides enough information to solve for the unknown using the fraction provided. Step 3: Combine the statements (if necessary) – If neither statement alone is sufficient, check if both statements together provide the information needed. It’s important to remember that in data sufficiency questions, you do not need to calculate the exact answer; you only need to decide whether the statements give you enough information to solve the problem. Example Problems with Solutions Here’s an example data sufficiency question involving fractions: “If 2/3 of a team are engineers, and Statement (1) says that the total number of team members is 15, while Statement (2) says that 10 team members are engineers, can you determine the number of non-engineers on the team?” Solution: By analyzing both statements, you can use Statement (1) to calculate that 2/3 of 15 members are engineers (i.e., 10 engineers), so 5 team members are not engineers. Therefore, each statement alone provides sufficient information to solve the problem. Common Mistakes Students Make with GMAT Fractions Fractions can be tricky to work with, and many students make common mistakes when dealing with them on the GMAT. Understanding these mistakes can help you avoid them and improve your score. Misinterpreting Fraction Problems A common mistake is misinterpreting fraction problems, particularly in word problems. Students often confuse the part with the whole or fail to properly understand what the fraction is representing. For example: “If 1/4 of a shipment is damaged, how much is undamaged?” Many students mistakenly calculate 1/4 of the shipment instead of subtracting 1/4 from the whole, which leaves 3/4 of the shipment undamaged. Not Simplifying Fractions Another common mistake is failing to simplify fractions when it is appropriate. Simplified fractions make calculations easier, and neglecting to simplify can lead to errors in solving GMAT problems. For example, when solving (6/8) (4/6), it’s better to simplify 6/8 to 3/4 and 4/6 to 2/3 before multiplying, which makes the problem easier to manage and reduces the chance of mistakes. Strategies to Avoid These Mistakes Here are some strategies to help avoid these common mistakes: Always read the problem carefully to understand what the fraction represents. Simplify fractions whenever possible before performing any calculations. Double-check your interpretation of the problem to ensure you are calculating the correct value. Practice fraction-based GMAT problems to become more comfortable with how they are tested on the exam. Additional Practice Questions Here are a few additional practice questions to help reinforce your understanding of fractions on the GMAT: If 3/5 of a group of 20 people are men, how many women are in the group? If a recipe calls for 1/3 cup of sugar and you want to double the recipe, how much sugar do you need? If 7/10 of a class passed an exam, how many students failed if the class has 40 students? Practice Questions for GMAT Fractions Practicing GMAT fraction questions is one of the most effective ways to improve your understanding of how fractions are tested. These questions often require quick calculations, conversions, and simplifications. Below, you’ll find a variety of practice questions designed to test your ability to work with fractions under different problem types. Fraction-Based Problem-Solving Questions These questions involve applying your knowledge of fractions to solve real GMAT-style problems. Try to work through them without using a calculator, as this will improve your mental math skills for the exam. Question 1: If 2/3 of a box is filled with marbles and the total number of marbles is 60, how many marbles are in the box? Question 2: A team’s winning percentage is 75%. What fraction of their games have they won? Question 3: If 1/5 of the employees at a company are managers, and the company has 200 employees, how many managers are there? Question 4: A recipe requires 3/4 cup of flour. If you want to make 1/2 of the recipe, how much flour should you use? Question 5: If a tank is 2/3 full and holds 90 liters of water when full, how many liters are in the tank? Solutions to Practice Problems Here are the solutions to the practice problems. Compare your answers to these solutions to see where you may have made mistakes or to reinforce your understanding of the concepts. Solution 1: To find 2/3 of 60, multiply 60 by 2/3. 60 (2/3) = 40 marbles. Solution 2: A winning percentage of 75% is the same as 3/4. The fraction of games won is 3/4. Solution 3: 1/5 of 200 is 40. So, there are 40 managers. Solution 4: Half of 3/4 is 3/8, so you will need 3/8 cup of flour. Solution 5: 2/3 of 90 liters is 60 liters. The tank contains 60 liters of water. These types of problems help you practice real GMAT scenarios involving fractions. The more comfortable you become with solving these problems, the faster and more accurate your calculations will be on the actual exam. GMAT Fractions in Word Problems Fractions are commonly found in GMAT word problems. These problems require you to translate real-life scenarios into mathematical terms using fractions. Whether you’re working with percentages, ratios, or quantities, being able to manipulate fractions is essential for solving word problems effectively. How Fractions Are Used in Word Problems In GMAT word problems, fractions are often used to represent parts of a whole, percentages, or proportions. For instance, a problem might ask how much of a total budget is allocated to a specific department or what fraction of a total group is involved in a task. Here’s an example: “A class of 80 students went on a field trip. If 1/4 of the students missed the bus, how many students missed the bus?” In this problem, you need to find 1/4 of 80. Multiply 80 by 1/4 to get 20 students who missed the bus. Step-by-Step Approach to Solving Word Problems Involving Fractions Follow these steps to solve word problems that involve fractions: Step 1: Read the problem carefully – Understand what part of the whole the fraction is representing. Step 2: Identify the total – Find the total quantity or value that the fraction is being applied to. Step 3: Multiply the total by the fraction – This will give you the part of the total that the fraction represents. Step 4: Double-check your work – Ensure that your interpretation of the fraction is correct and that your calculations are accurate. Let’s go through another example: “A company has 120 employees. If 3/5 of the employees work in the sales department, how many employees work in sales?” To solve this, multiply 120 by 3/5. The calculation is: 120 (3/5) = 72 employees work in sales. Common Types of Word Problems Using Fractions Here are the most common types of word problems on the GMAT that use fractions: Percentage Problems: These problems often involve converting percentages to fractions. For example, 25% of a group can be expressed as 1/4 of the group. Ratio Problems: Ratios like 2:3 are expressed as fractions (2/5 and 3/5) to find proportions of a total. Proportion Problems: These involve direct application of fractions to find parts of a whole, such as calculating 2/3 of a budget or 1/4 of a shipment. Being able to identify these types of problems and apply fractions correctly is essential for success on the GMAT. GMAT Fractions in Ratio and Proportion Problems Ratio and proportion problems are commonly tested on the GMAT, and many of these involve fractions. In these types of problems, you are often asked to find the relationship between two or more quantities, which can usually be expressed as fractions. Understanding how to convert ratios into fractions and apply them correctly is essential for solving these questions. How Ratios and Proportions Are Tested Using Fractions In GMAT ratio and proportion problems, fractions help to express the relative sizes of two or more quantities. For example, a ratio of 3:2 means that for every 5 units, 3 belong to one group, and 2 belong to the other. This ratio can be expressed as fractions: 3/5 for the first group and 2/5 for the second group. Let’s look at an example: “A company has a ratio of 3 salespeople for every 2 marketing personnel. If there are 45 total employees, how many are salespeople?” To solve this, you first add the ratio parts (3 + 2 = 5). This means salespeople represent 3/5 of the total employees. So, multiply 45 by 3/5 to find the number of salespeople: 45 (3/5) = 27 salespeople Practice Ratio and Proportion Problems Below are more practice problems for ratio and proportion involving fractions: Question 1: In a class, the ratio of boys to girls is 3:2. If there are 30 students in the class, how many are boys? Question 2: A recipe requires a ratio of 2 parts sugar to 3 parts flour. If you have 10 cups of flour, how much sugar should you use? Question 3: A company’s staff ratio is 5 developers to 3 designers. If there are 64 total staff members, how many developers are there? Question 4: A juice blend has a ratio of 4 parts apple juice to 1 part orange juice. If there are 25 liters of the blend, how many liters of apple juice are in the blend? Question 5: In a school, the ratio of teachers to students is 1:20. If there are 300 students, how many teachers are there? Solutions to Ratio and Proportion Problems Solution 1: The ratio of boys to total students is 3/5. Multiply 30 by 3/5 to get 18 boys. Solution 2: The ratio of sugar to flour is 2/3. Multiply 10 by 2/3 to get 6.67 cups of sugar. Solution 3: The ratio of developers to total staff is 5/8. Multiply 64 by 5/8 to get 40 developers. Solution 4: The ratio of apple juice to total blend is 4/5. Multiply 25 by 4/5 to get 20 liters of apple juice. Solution 5: The ratio of teachers to students is 1/20. Multiply 300 by 1/20 to get 15 teachers. Comparing Fractions on the GMAT Comparing fractions is another essential skill for the GMAT. In some questions, you may need to determine which of two fractions is larger or smaller. This is especially common in data sufficiency and problem-solving questions. Knowing how to quickly compare fractions can save valuable time during the test. We can help you get into your Dream Schools If you book a call, we will cover the following: Detailed Analysis of Profile Strengths and Weaknesses ⁠Ideal 8 Schools to target What GMAT Score do you really need ⁠Things you can do now to Improve your Profile How to improve your Admit and Scholarship chance Name Enter your Name Email ID Enter your Email Contact Number +91 Enter your Number Book a Free Session Now Your Appointment is being confirmed! (Step 1 of 2) Just give us one last detail and you are good to go! Which intake are you applying to? (This would be the year you will be starting your studies in) Please Choose Atleast One Option 2026 2027 and Later Submit Methods for Comparing Fractions There are a few methods you can use to compare fractions: Cross-Multiplication: Multiply the numerator of the first fraction by the denominator of the second, and the numerator of the second by the denominator of the first. The larger product will indicate the larger fraction. Converting to Decimals: Convert both fractions to decimal form and compare their values. This method works best when dealing with fractions that are commonly used. Benchmark Fractions: Compare each fraction to a known benchmark, like 1/2 or 1. For example, fractions larger than 1/2 are always greater than those less than 1/2. Let’s look at an example: “Which is larger, 5/8 or 7/12?” Using cross-multiplication: 5 12 = 60 8 7 = 56 Since 60 is greater than 56, 5/8 is larger than 7/12. Practice Questions for Comparing Fractions Here are several questions to help you practice comparing fractions: Question 1: Which is larger, 3/4 or 5/6? Question 2: Compare 7/10 and 2/3. Which fraction is greater? Question 3: Is 5/12 greater than or less than 1/3? Question 4: Which is smaller, 9/16 or 1/2? Question 5: Compare 4/5 and 9/10. Which is greater? Solutions to Comparing Fractions Questions Solution 1: Using cross-multiplication: 3 6 = 18, 4 5 = 20. Since 20 is larger, 5/6 is greater than 3/4. Solution 2: Convert to decimals: 7/10 = 0.7, 2/3 = 0.666... So, 7/10 is larger. Solution 3: Using cross-multiplication: 5 3 = 15, 12 1 = 12. Since 15 is larger, 5/12 is greater than 1/3. Solution 4: Convert to decimals: 9/16 = 0.5625, 1/2 = 0.5. So, 1/2 is smaller than 9/16. Solution 5: Convert to decimals: 4/5 = 0.8, 9/10 = 0.9. So, 9/10 is greater than 4/5. Related Blogs GMAT arithmetic tricks Coordinate geometry GMAT GMAT algebra practice questions Time and work problems GMAT GMAT probability cheat sheet Conclusion Understanding and mastering GMAT fractions is essential for improving your performance in the quantitative section. By learning how to convert, compare, and simplify fractions, you’ll be able to solve problems faster and with greater accuracy. Regular practice with real GMAT-style questions, focusing on fractions in ratios, proportions, and word problems, will help you gain confidence in handling these concepts. Keep practicing, and with time, fractions will become one of the most manageable parts of the GMAT. Know Your Author View Profile Abhyank Srinet | Study Abroad Expert Abhyank Srinet, the founder of MiM-Essay, is a globally recognized expert in study abroad and admission consulting. His passion is helping students navigate the complex world of admissions and achieve their academic dreams. Abhyank earned a Master's degree in Management from ESCP Europe, where he developed his skills in data-driven marketing strategies, driving growth in some of the most competitive industries....Read full Bio Abhyank has helped over 10,000+ students get into top business schools with a 98% success rate over the last seven years. He and his team offer thorough research, careful shortlisting, and efficient application management from a single platform. His dedication to education also led him to create MentR-Me, an AI-powered platform that offers personalized guidance and resources, including profile evaluation, application assistance, and mentoring from alumni of top global institutions. Continuously adopting the latest strategies, Abhyank is committed to ensuring that his clients receive the most effective guidance. His profound insights, extensive experience, and unwavering dedication have helped his clients securing of over 100 crores in scholarships, making him an invaluable asset for individuals aiming to advance their education and careers and leading both his ventures to seven-figure revenues. Want to 3X your Admission Chances to the Best Schools? Connect with our Experts and get: Detailed Profile Analysis Personalised School Suggestions Access to Application Dashboard Book your FREE Session Now! Related Blogs Carnegie Mellon University MS in Data science Revi Launch Your Career: Top Business Schools in Austra Canada Student Visa Process: How to Apply and Requ How to Write a Winning SOP for MIM (Master’s in Ma You may also like these Blogs Study Abroad Carnegie Mellon University MS in Data science Review 8 minutes Explore Carnegie Mellon University's MS in Data Science program. Learn about its top rankings, fees, curriculum, and strong career opportunities with high average starting salaries. Australia Study Abroad Launch Your Career: Top Business Schools in Australia 10 minutes Australia is home to some of the world's top business schools, offering exceptional education, strong global networks, and a culturally diverse learning environment. Discover the opportunities and advantages that these esteemed institutions provide, and find the perfect place to advance your busines Canada Study Abroad Canada Student Visa Process: How to Apply and Requirements 10 minutes As a student, visa process could be difficult. But all details regarding visa process for Canada Student visa is shared here. Study Abroad How to Write a Winning SOP for MIM (Master’s in Management) 7 minutes Discover how to write a winning MiM SOP for MiM with expert tips, ideal format, and real samples. Perfect your SOP and boost your chances for top B-schools. Start now. 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https://miamioh.ecampus.com/fishmans-pulmonary-diseases-disorders/bk/9780071807289?srsltid=AfmBOoriyaXNlpLmewl7_Y53mkURJVwHIsYDk2jrnkF-FICG-VRclqzj
Fishman's Pulmonary Diseases and Disorders, 2-Volume Set, 5th edition Skip Navigation MIAMI UNIVERSITY OFFICIAL BOOKSTORE Login/Sign Up Home Miami Alumni Miami Athletics Shopping Cart (0) Shop MENU Shop Textbooks Campus Locations Login/Sign Up Back| Campus Locations Home Home Miami Alumni Miami Alumni Miami Athletics Miami Athletics Back| Shop Clothing Clothing Accessories Accessories Gifts Gifts Graduation Graduation Supplies Supplies Back| Clothing Kids Men Sweatshirts Women View All Back| Accessories For You For Your Car For Your Home For Your Pet For Your Tech View All Back| Gifts Artwork Cooking Essentials Games Gift Wraps Holiday Home Decor Mascot Office Decor Outdoor/Recreation View All Back| Graduation Graduation Gear Graduation Gifts View All Back| Supplies Art Supplies For Your Office Medical Supplies Office Supplies School Supplies View All Kids Back| Kids Bibs Bottoms Dresses Headwear Hoodies Matching Sets Onesies Shirts Sweatshirts Men Back| Men Bottoms Footwear Hoodies Jerseys Mens Apparel Outerwear Polos Shirts Sweatshirts T-Shirts Sweatshirts Back| Sweatshirts Women Back| Women Bottoms Dresses Headwear Hoodies Outerwear Pants Shirts Sweatshirts Undergarments View All Clothing > Miami Merger Myaamia Heritage Collection Alumni Collection Miami Regionals Custom Items Miami Cradle of Coaches collection For You Back| For You Backpacks Bags Buttons Drinkware Fan Gear Flags Hair Accessories Headwear ID Holders Jewelry Keychains Knitwear Lanyards Lapel Pins Pennants Socks Ties Umbrellas Wallets For Your Car Back| For Your Car Decals License Plates For Your Home Back| For Your Home Banners Blankets Cooking Essentials Decals Drinkware Flags Frames Home Decor Magnets Pillows Signs Stickers For Your Pet Back| For Your Pet Beds Bowls Charms Clothes Toys For Your Tech Back| For Your Tech Computer Accessories Phone Accessories View All Accessories > Artwork Back| Artwork Ornaments Wall Art Cooking Essentials Back| Cooking Essentials Food Games Back| Games Balls Puzzles Gift Wraps Back| Gift Wraps Gift Bags Holiday Back| Holiday Ornaments Stocking Home Decor Back| Home Decor Candles Mascot Back| Mascot Plushies Office Decor Back| Office Decor Desk Accessories Outdoor/Recreation Back| Outdoor/Recreation Tailgate View All Gifts > Graduation Gear Back| Graduation Gear Caps and Gowns Hoods Stoles Study Abroad Sashes Tassel Tassels Graduation Gifts Back| Graduation Gifts Diploma Frames Yard Signs View All Graduation > Art Supplies Back| Art Supplies Art Supply Products For Your Office Back| For Your Office Desk Accessories Pens Medical Supplies Back| Medical Supplies Kits Office Supplies Back| Office Supplies Padfolios School Supplies Back| School Supplies Folders Notebooks Planners View All Supplies > Erin Condren Textbooks Search Shopping Cart (0) Write a Review Fishman's Pulmonary Diseases and Disorders, 2-Volume Set, 5th edition byGrippi, Michael; Elias, Jack; Fishman, Jay; Pack, Allan; Senior, Robert; Kotloff, Robert Edition: 5th ISBN13: 9780071807289 ISBN10: 0071807284 Format: Hardcover Pub. Date: 2015-04-14 Publisher(s): McGraw Hill / Medical Other versions by this Author This Item Qualifies for Free Shipping! Excludes marketplace orders. List Price: ~~$472.50~~ Rent Textbook Select for Price Add to Cart There was a problem. Please try again later. Buy New Arriving Soon. Will ship when available. $450.00 Add to Cart Rent Digital Rent Digital Options Online:1825 Days access, Downloadable:Lifetime Access - $506.25 Online:1825 Days access Downloadable:Lifetime Access $506.25 $506.25 Add to Cart Used Textbook We're Sorry Sold Out Buy from our Marketplace starting at $75.36 How Marketplace Works: This item is offered by an independent seller and not shipped from our warehouse Item details like edition and cover design may differ from our description; see seller's comments before ordering. Sellers much confirm and ship within two business days; otherwise, the order will be cancelled and refunded. Marketplace purchases cannot be returned to eCampus.com. Contact the seller directly for inquiries; if no response within two days, contact customer service. Additional shipping costs apply to Marketplace purchases. Review shipping costs at checkout. Currently unavailable Preferred Sellers are consistently the most reliable sellers on our Marketplace with a track record of providing great service to their customers. We regularly review these sellers to ensure that you can trust them. Summary Turn to the field's definitive text for a thorough understanding of the clinical and scientific aspects of pulmonary medicine Since 1980, Fishman's Pulmonary Diseases and Disorders has delivered unparalleled coverage of pulmonary medicine and the underlying basic and applied science upon which clinical practice is based. The Fifth Edition, with 270 contributing authors, includes over 2,000 illustrations, 60 videos, and 18,000 references. The book opens with a comprehensive overview of the scientific basis of lung function in health and disease. It then provides detailed coverage of the broad array of diseases and disorders affecting the respiratory system, including obstructive and restrictive diseases, pulmonary vascular disorders, sleep-disordered breathing, lung neoplasms, respiratory infections, and respiratory failure, among others. The Fifth Edition has been completely updated to reflect the many advancements that have been made in pulmonary medicine over the past few years, including: Molecular development of the lung Stem cells and respiratory disease Genetics of pulmonary disease and the growth of personalized medicine Technical advances in lung transplantation Growth in immunology and immunosuppressive management Diagnosis and treatment of pulmonary hypertension Circadian rhythms and sleep biology Rapid evolution in lung imaging techniques, including functional imaging Contemporary interventional bronchoscopic techniques You will also find state-of-the-art coverage of the latest topics in critical care medicine, including: Early diagnosis and management of sepsis Multiple organ dysfunction syndrome (MODS) Acute respiratory distress syndrome (ARDS) Management of agitation and delirium in the ICU The newly defined entity of "chronic critical illness" We are currently experiencing difficulties. 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https://www.scribbr.com/statistics/mode/
### Have a language expert improve your writing Proofreading Services ### Run a free plagiarism check in 10 minutes Plagiarism Checker ### Generate accurate citations for free Citation Generator Home Knowledge Base Statistics How to Find the Mode | Definition, Examples & Calculator How to Find the Mode | Definition, Examples & Calculator Published on October 2, 2020 by Pritha Bhandari. Revised on June 21, 2023. The mode or modal value of a data set is the most frequently occurring value. It’s a measure of central tendency that tells you the most popular choice or most common characteristic of your sample. When reporting descriptive statistics, measures of central tendency help you find the middle or the average of your data set. The three most common measures of central tendency are the mode, median, and mean. Table of contents How many modes can you have? Mode calculator Find the mode (by hand) Find the mode with grouped data When to use the mode Other interesting articles Frequently asked questions about the mode How many modes can you have? A data set can often have no mode, one mode or more than one mode – it all depends on how many different values repeat most frequently. Your data can be: without any mode unimodal, with one mode, bimodal, with two modes, trimodal, with three modes, or multimodal, with four or more modes. Prevent plagiarism. Run a free check. Try for free Mode calculator You can calculate the mode by hand or with the help of our mode calculator below. Find the mode (by hand) To find the mode, follow these two steps: If the data for your variable takes the form of numerical values, order the values from low to high. If it takes the form of categories or groupings, sort the values by group, in any order. Identify the value or values that occur most frequently. Numerical mode example Your data set is the ages of 6 college students. Data set | | | | | | | | --- --- --- | Participant | A | B | C | D | E | F | | Age | 19 | 22 | 20 | 21 | 22 | 23 | By ordering the values from low to high, we can easily see the value that occurs most frequently. Ordered data set | | | | | | | | --- --- --- | Age | 19 | 20 | 21 | 22 | 22 | 23 | The mode of this data set is 22. Categorical mode example Your data set contains the highest education levels of the participants’ parents. Data set | | | | | | | | --- --- --- | Participant | A | B | C | D | E | F | | Parents’ education level | Bachelor’s degree | Master’s degree | High school diploma | Bachelor’s degree | Doctoral degree | Master’s degree | To sort the values by group, you create a simple frequency table. Place the categories on the left hand side and the frequencies on the right hand side. Frequency table | Parents’ education level | Frequency | --- | | Bachelor’s degree | 2 | | Master’s degree | 2 | | High school diploma | 1 | | Doctoral degree | 1 | From the table, you can see that there are two modes. This means you have a bimodal data set. The modes are Bachelor’s degree and Master’s degree. Find the mode with grouped data A grouped frequency table organizes large numerical data sets into intervals or classes of values and reports the frequency of values in each class. For grouped data, you can report the mode in two ways: the modal class is the grouping with the highest frequency of values. the modal value is estimated as the midpoint of the modal class. The mode is only an estimate in this case, because the actual values within the modal class are unknown. Modal class and modal value example You have a data set that includes the average reaction times of participants. You organize the data into a frequency table. Reaction times are placed in classes of 100 milliseconds each. The frequency column shows the number of participants within each class. Grouped frequency table | Reaction time (milliseconds) | Frequency | --- | | 200–299 | 6 | | 300–399 | 13 | | 400–499 | 17 | | 500–599 | 25 | | 600–699 | 21 | | 700–799 | 12 | | 800–899 | 4 | You can visualize your data set by plotting your data on a histogram. The mode is the value with the highest peak on a histogram or bar chart. From your table or histogram, you can see that the modal class – the group in which values appear most frequently – is 500–599 milliseconds. Therefore, the mode is estimated to be at the midpoint of this class: 550 milliseconds. Importantly, the choice of intervals in grouped data can have a large impact on the mode. For example, changing the intervals from 100 ms long to 50 or 200 ms long could result in completely different modes. Prevent plagiarism. Run a free check. Try for free When to use the mode The level of measurement of your variables determines when you should use the mode. The mode works best with categorical data. It is the only measure of central tendency for nominal variables, where it can reflect the most commonly found characteristic (e.g., demographic information). The mode is also useful with ordinal variables – for example, to reflect the most popular answer on a ranked scale (e.g., level of agreement). For quantitative data, such as reaction time or height, the mode may not be a helpful measure of central tendency. That’s because there are often many more possible values for quantitative data than there are for categorical data, so it’s unlikely for values to repeat. Example of quantitative data with no mode You collect data on reaction times in a computer task, and your data set contains values that are all different from each other. Data set with no mode | | | | | | | | | | | --- --- --- --- --- | | Reaction time (milliseconds) | 267 | 345 | 421 | 324 | 401 | 312 | 382 | 298 | 303 | In this data set, there is no mode, because each value occurs only once. Other interesting articles If you want to know more about statistics, methodology, or research bias, make sure to check out some of our other articles with explanations and examples. Statistics Statistical power Pearson correlation Variance Degrees of freedom Statistical significance Skewness Methodology Cluster sampling Stratified sampling Focus group Systematic review Ethnography Double-Barreled Question Research bias Implicit bias Publication bias Cognitive bias Placebo effect Pygmalion effect Hindsight bias Overconfidence bias Frequently asked questions about the mode How do I find the mode? : To find the mode: If your data is numerical or quantitative, order the values from low to high. If it is categorical, sort the values by group, in any order. Then you simply need to identify the most frequently occurring value. Can there be more than one mode? : A data set can often have no mode, one mode or more than one mode – it all depends on how many different values repeat most frequently. Your data can be: without any mode unimodal, with one mode, bimodal, with two modes, trimodal, with three modes, or multimodal, with four or more modes. What are measures of central tendency? : Measures of central tendency help you find the middle, or the average, of a data set. The 3 most common measures of central tendency are the mean, median and mode. The mode is the most frequent value. The median is the middle number in an ordered data set. The mean is the sum of all values divided by the total number of values. Which measures of central tendency can I use? : The measures of central tendency you can use depends on the level of measurement of your data. For a nominal level, you can only use the mode to find the most frequent value. For an ordinal level or ranked data, you can also use the median to find the value in the middle of your data set. For interval or ratio levels, in addition to the mode and median, you can use the mean to find the average value. Cite this Scribbr article If you want to cite this source, you can copy and paste the citation or click the “Cite this Scribbr article” button to automatically add the citation to our free Citation Generator. Is this article helpful? You have already voted. Thanks :-) Your vote is saved :-) Processing your vote... Pritha Bhandari Pritha has an academic background in English, psychology and cognitive neuroscience. As an interdisciplinary researcher, she enjoys writing articles explaining tricky research concepts for students and academics. Why do I see ads? Ads help us keep our tools free for everyone. Scribbr customers enjoy an ad-free experience! Other students also liked #### Central Tendency | Understanding the Mean, Median & Mode Measures of central tendency help you find the middle, or average, of a data set. Mean, median and mode are the 3 main measures. #### Levels of Measurement | Nominal, Ordinal, Interval and Ratio Levels of measurement tell you how precisely variables are recorded. The level of measurement determines how you can analyze your data. #### Descriptive Statistics | Definitions, Types, Examples Descriptive statistics summarize the characteristics of a data set. There are three types: distribution, central tendency, and variability. What is your plagiarism score? Scribbr Plagiarism Checker
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In the Light of Evolution: Volume III: Two Centuries of Darwin (2009) Chapter: 13 Darwin and the Scientific Method--Francisco J. Ayala Get This Book Visit NAP.edu/10766 to get more information about this book, to buy it in print, or to download it as a free PDF. « Previous: Part IV: THE DARWINIAN LEGACY, 150 YEARS LATER Page 267 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. 13 Darwin and the Scientific Method FRANCISCO J. AYALA There is a contradiction between Darwin’s methodology and how he described it for public consumption. Darwin claimed that he proceeded “on true Baconian [inductive] principles and without any theory collected facts on a wholesale scale.” He also wrote, “How odd it is that anyone should not see that all observation must be for or against some view if it is to be of any service!” The scientific method includes 2 episodes. The first consists of formulating hypotheses; the second consists of experimentally testing them. What differentiates science from other knowledge is the second episode: subjecting hypotheses to empirical testing by observing whether or not predictions derived from a hypothesis are the case in relevant observations and experiments. A hypothesis is scientific only if it is consistent with some but not other possible states of affairs not yet observed, so that it is subject to the possibility of falsification by reference to experience. Darwin occupies an exalted place in the history of Western thought, deservedly receiving credit for the theory of evolution. In The Origin of Species, he laid out the evidence demonstrating the evolution of organisms. More important yet is that he discovered natural selection, the process that accounts for the adaptations of organisms and their complexity and diversification. Natural selection and other causal processes of evolution are investigated by formulating and testing hypotheses. Darwin advanced hypotheses in multiple fields, Department of Ecology and Evolutionary Biology, University of California, Irvine, CA 92697. Page 268 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × including geology, plant morphology and physiology, psychology, and evolution, and subjected them to severe empirical tests. DARWIN AND THE PHILOSOPHERS There is an apparent contradiction between how Darwin (Fig. 13.1) proceeded in his scientific research and how he described it for public consumption, between what he said in his published writings about his scientific methodology and what he wrote in his notebooks, correspondence, and autobiography. The opening paragraph of The Origin of Species (Fig. 13.2) reads as follows: When on board H.M.S. Beagle, as naturalist, I was much struck with certain facts in the distribution of the inhabitants of South America, and in the geological relations of the present to the past inhabitants of that continent. These facts seemed to me to throw some light on the origin of species—that mystery of mysteries, as it has been called by one of our greatest philosophers. On my return home, it occurred to me, in 1837, that something might perhaps be made out on this question by patiently accumulating and reflecting on all sorts of facts which could possibly have any bearing on it. After 5 years’ work I allowed myself to speculate on the subject, and drew up some short notes; these I enlarged in 1844 into a sketch of the conclusions, which then seemed to me probable: from that period to the present day I have steadily pursued the same object. Darwin claims to have followed the inductionist canon prevalent among British contemporary philosophers and economists, such as John Stuart Mill (1843), and earlier authorities, notably the statesman and philosopher, Francis Bacon in his Novum Organum (Anderson, 1960). The inductionist canon called for making observations without prejudice as to what they might mean and accumulating observations related to a particular subject so that a universal statement or conclusion could eventually emerge from them. Indeed, in one place in his Autobiography, Darwin affirms that he proceeded “on true Baconian principles and without any theory collected facts on a wholesale scale” (Barlow, 1958, p. 119). The facts are very different from these claims, however. Darwin’s notebooks and private correspondence show that he entertained the hypothesis of the evolutionary transmutation of species shortly after returning from the voyage of the Beagle and, all important, that the hypothesis of natural selection occurred to him in 1838; several years before he claims to have allowed himself for the first time “to speculate on the subject.” Between the return of the Beagle on October 2, 1836, and publication of Page 269 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × FIGURE 13.1 Charles Darwin, circa 1854 (courtesy of Professor G. Evelyn Hutchison). Origin of Species in 1859 (Darwin, 1859, 6th ed.) (and, indeed, until the end of his life), Darwin relentlessly pursued empirical evidence to corroborate the evolutionary origin of organisms and to test his theory of natural selection, which he saw as the explanatory process accounting for the adaptive organization of living beings and their diversification and change through time. Why this disparity between what Darwin was doing and what he claimed? There are at least 2 reasons. First, in the temper of the times, “hypothesis” was a term often reserved for metaphysical speculations without empirical substance. This is the reason Newton, the greatest-ever theorist among scientists, had also claimed, hypotheses non fingo (“I fabricate no hypotheses”). Darwin expressed distaste and even contempt Page 270 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × FIGURE 13.2 Title page of The Origin of Species. for empirically untestable hypotheses. He wrote of Herbert Spencer: “His deductive manner of treating any subject is wholly opposed to my frame of mind. His conclusions never convince me. His fundamental generalizations (which have been compared in importance by some persons with Newton’s Laws!), which I daresay may be very valuable under a philosophical point of view, are of such a nature that they do not seem to me to be of any strictly scientific use. They partake more of the nature of definitions than of laws of nature. They do not aid me in predicting what will happen in any particular case” (Barlow, 1958, p. 109). There is another reason, a tactical one, Darwin claimed to proceed according to inductive canons: he did not want to be accused of subjective Page 271 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × bias in the evaluation of empirical evidence. Darwin’s true colors are shown in a letter to a young scientist written in 1863: “I would suggest to you the advantage, at present, of being very sparing in introducing theory in your papers (I formerly erred much in Geology in that way); let theory guide your observations, but till your reputation is well established, be sparing of publishing theory. It makes persons doubt your observations” (Darwin, 1903, Vol. 2, p. 323). Nowadays also, scientists, young or old, often report their work so as to make their hypothesis appear as afterthoughts, conclusions derived from the observations or experiments made, rather than as preconceptions tested by empirical observations designed precisely, as it is most often the case in many scientific disciplines, for the purpose of testing a particular “preconception,” a hypothesis. Nevertheless, it is becoming more and more the case that experiments and observations are planned and reported as specific tests of a particular hypothesis. “Let theory guide your observations.” Indeed, Darwin had no use for the empiricist claim that a scientist should not have a preconception or hypothesis that would guide his work. Otherwise, as he wrote, one “might as well go into a gravel pit and count the pebbles and describe the colors. How odd it is that anyone should not see that observation must be for or against some view if it is to be of any service” (Darwin, 1903, Vol. 1, p. 195). He acknowledged the heuristic role of hypotheses, which guide empirical research by suggesting what is worth observing, what evidence to seek. In his Autobiography, he acknowledges that “I cannot avoid forming one [hypothesis] on every subject” (Barlow, 1958, p. 141). Darwin advanced hypotheses in multiple fields, including geology, plant morphology and physiology, psychology, and evolution, and subjected his hypotheses to severe empirical tests. Herein lies the solution to the historical conundrum, often noted by historians and philosophers, that he delayed for 2 decades publication of his theory of natural selection as an explanation for the adaptations and diversification of organisms, which he had discovered in 1838, but did not publish until 1859, in Origin. (The delay might have been longer were it not for Wallace’s letter of 1858 announcing his independent discovery of natural selection.) Darwin was aware of the major implications of his theory, namely, bringing the adaptations and diversity of organisms into the realm of science rather than being accounted for by direct creation, as was generally accepted at the time (Ayala, 2007). He spent many years testing his theory of natural selection with observations and experiments that seemed likely to contradict his theory, if it were not correct. Historians have often thought that his 4 volumes on barnacles, living and fossils (Darwin, 1851a,b, 1854a,b) and his studies on the fertilization of orchids (Darwin, 1862), and others, were distractions. They were not distractions, but rather severe tests of his theory of natural selection. Page 272 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × INDUCTION AND EMPIRICISM It is a common misconception, shared by many scientists, that science proceeds by “accumulating experimental facts and drawing up a theory from them,” as François Jacob (1988, pp. 224–225) had believed when he started the research on bacteriophage replication that would lead to his receiving, in 1965, the Nobel Prize for physiology or medicine. This misconception is encased in the much repeated assertion that science is inductive, a notion that can be traced to the English statesman and essayist Francis Bacon (1561–1626). Bacon had an influential role in shaping modern science by his criticism of the prevailing metaphysical speculations of medieval scholastic philosophers. In the 19th century the most articulate proponent of inductivism was John Stuart Mill (1806–1873). Induction was proposed by Bacon and Mill as a method of achieving objectivity while avoiding subjective preconceptions and obtaining empirical rather than abstract or metaphysical knowledge. In its extreme form the inductivist canon would hold that a scientist should observe any phenomena that he encounters in his experience and record them without any preconceptions as to what to observe or what the truth about his observations might be. Truths of universal validity would be expected eventually to emerge, as a result of the relentless accumulation of unprejudged observations. The methodology proposed may be trivially exemplified as follows. A scientist measuring and recording everything that confronts him observes a tree with leaves. A second tree, and a third, and many others, are all observed to have leaves. Eventually, he formulates a universal statement, “all trees have leaves.” This inductive process fails to account for the actual methodology of science. First of all, no scientist works without any preconceived plan as to what kind of phenomena to observe. Scientists choose for study objects or events that, in their opinion, are likely to provide answers to questions that interest them. Otherwise, as Darwin wrote, “one might as well go into a gravel pit and count the pebbles and describe the colors.” A scientist whose goal was to record carefully every event observed in all waking moments of his life would not contribute much to the advance of science; more likely than not, he might be considered mad by his colleagues. Moreover, induction fails to arrive at universal truths. No matter how many singular statements may be accumulated, no universal statement can be logically justified by such an accumulation of observations. Even if all trees so far observed have leaves, or all swans observed are white, it remains a logical possibility that the next tree will not have leaves, or the next swan will not be white. The step from numerous singular statements to a universal one involves logical amplification. The universal statement has greater logical content (it says more) than the sum of all singular statements. Page 273 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Another serious logical difficulty with the proposal that induction is the method of science, is that scientific hypotheses and theories are formulated in abstract terms that do not occur at all in the description of empirical events. Mendel, the founder of genetics, observed in the progeny of hybrid plants that alternative traits segregated according to certain proportions. Repeated observations of these proportions could never have led inductively to the formulation of his hypothesis that “factors” (genes) exist in the sex cells and are rearranged in the progeny according to certain rules. The genes were not observed and thus could not be included in statements reflecting what Mendel observed. Natural selection, like gravity or electricity, is not directly observed by a simple examination of nature at a particular time or place. The most interesting and fruitful scientific hypotheses are not simple generalizations. Instead, scientific hypotheses are creations of the mind, imaginative suggestions as to what might be true. Induction fails in all 3 counts pointed out. It is not a method that ensures objectivity and avoids preconceptions, it is not a method to reach universal truths, and it is not a good description of the process by which scientists formulate hypotheses and other forms of scientific knowledge. It is a different matter that a scientist may come upon a new idea or develop a hypothesis as a consequence of repeated observation of phenomena that might be similar or share certain traits. But how we come upon a new idea is quite a different matter from how is it that we come to accept something as established scientific knowledge. THE HYPOTHETICO-DEDUCTIVE METHOD New ideas in science are advanced in the form of conjectures or hypotheses, which may be more or less precisely formulated and be of lesser or greater generality. However, it is essential to the scientific process that any hypothesis be “tested” by reference to the natural world that we experience with our senses. The tests to which scientific ideas are subjected include contrasting any hypothesis with the world of experience in a manner that must leave open the possibility that one might reject a particular hypothesis if it leads to wrong predictions about the world of experience. The possibility of empirical falsification of a hypothesis is carried out by ascertaining whether or not precise predictions derived as logical consequences from the hypothesis agree with the state of affairs found in the empirical world. A hypothesis that cannot be subject to the possibility of rejection by observation and experiment cannot be regarded as scientific. There are 2 basic components in the process by which scientific knowledge advances. The first component consists of the formulation of a con- Page 274 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × jecture or hypothesis about the natural world. The second component consists of testing the hypothesis by ascertaining whether deductions derived from the hypothesis are indeed the case in the real world. This procedural practice has become known as the hypothetico-deductive method, often characterized as “the” scientific method. It is of the essence of the testing process that the predictions derived from the hypothesis to be tested not be already known, if the observations to be made are to serve as a genuine test of the hypothesis. If a hypothesis is formulated to account for some known phenomena, these phenomena may provide credibility to the hypothesis, but by themselves do not amount to a genuine empirical test of it for the purpose of validating it. The value of a test increases to the extent that the predicted consequences appear to be more and more unlikely before the observations are made. The analysis of the hypothetico-deductive method may be traced to William Whewell (1794–1866) and William Stanley Jevons (1835–1882) in England and Charles S. Peirce (1838–1914) in the United States. In the 20th century, 2 philosophers who greatly contributed to identify the key features of the hypothetico-deductive method, and are broadly credited for this work, are Karl Popper (1902–1994) (Popper, 1959, 1963) and C. G. Hempel (1905–1997) (Hempel, 1965). But there is no better way of understanding the basic components of the scientific method, and its variations in different disciplines and peculiarities in different practitioners, than examining the work of great scientists, whose enormous accomplishments were made possible by their appropriate methodology. Early eminent practitioners of the hypothetico-deductive methodology include Blaise Pascal (1623–1662) and Isaac Newton (1624–1727). Among biologist contemporaries of Darwin, one might mention Claude Bernard (1813–1878), Louis Pasteur (1822–1895), and Gregor Mendel (1822–1884). IMAGINATION AND CORROBORATION Some of these scientists explicitly described the methodology they followed in their research. Notable is the case of Claude Bernard (1865), who clearly describes the 2 stages of the scientific method: formulation of a testable hypothesis and testing it. Moreover, Bernard explicitly asserts that scientific theories of necessity are only partial and provisional. A hypothesis is … the obligatory starting point of all experimental reasoning. Without it no investigation would be possible, and one would learn nothing: one could only pile up barren observations. To experiment without a preconceived idea is to wander aimlessly…. Those who have condemned the use of hypotheses and preconceived ideas in the experimental method have made the mistake of confusing the contriving of the experiment with the verification of its results…. When propounding a Page 275 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × general theory in science, the one thing one can be sure of is that, in the strict sense, such theories are mistaken. They are only partial and provisional truths which are necessary… to carry the investigation forward; they represent only the current state of our understanding and are bound to be modified by the growth of science. A contemporary scientist, the Nobel Prize recipient François Jacob, has described research in the lab as an interplay between imagination (hypothesis formulation) and experiment: What had made possible analysis of bacteriophage multiplication, and understanding of its different stages, was above all of the play of hypotheses and experiments, constructs of the imagination and inferences that could be drawn from them. Starting with a certain conception of the system, one designed an experiment to test one or another aspect of this conception. Depending on the results, one modified the conception to design another experiment. And so on and so forth. That is how research in biology worked. Contrary to what once I thought, scientific progress did not consist simply in observing, in accumulating experimental facts and drawing up a theory from them. It began with the invention of a possible world, or a fragment thereof, which was then compared by experimentation with the real world. And it was this constant dialogue between imagination and experiment that allowed one to form an increasingly fine-grained conception of what is called reality. Jacob (1988, pp. 224–225) As pointed out above, science is a complex enterprise that essentially consists of 2 interdependent episodes, one imaginative or creative, the other critical. To have an idea, advance a hypothesis, or suggest what might be true is a creative exercise. However, scientific conjectures or hypotheses must also be subject to critical examination and empirical testing. Scientific thinking may be characterized as a process of invention or discovery followed by validation or confirmation. One process concerns the formulation of new ideas (“acquisition of knowledge”), the other concerns their validation (“justification of knowledge”). Scientists like other people come upon new ideas in all sorts of ways: from conversation with other people, reading books and newspapers, inductive generalizations, and even dreams and mistaken observations. Newton is said to have been inspired by a falling apple. Kekulé had been unsuccessfully attempting to devise a model for the molecular structure of benzene. One evening he was dozing in front of the fire. The flames appeared to Kekulé as snake-like arrays of atoms. Suddenly one snake appeared to bite its own tail and then whirled mockingly in front of him. The circular appearance of the image inspired in him the model of benzene as a hexagonal ring. The model to explain the evolutionary diversification Page 276 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × of species came to Darwin while riding in his coach and observing the countryside. “I can remember the very spot in the road … when to my joy the solution came to me…. The solution, as I believe, is that the modified offspring … tend to become adapted to many and highly diversified places in the economy of nature” (Barlow, 1958, pp. 120–121). Hypotheses and other imaginative conjectures are the initial stage of scientific inquiry. It is the imaginative conjecture of what might be true that provides the incentive to seek the truth and a clue as to where we might find it. Hypotheses guide observation and experiment because they suggest what to observe. The empirical work of scientists is guided by hypotheses, whether explicitly formulated or simply in the form of vague conjectures or hunches about what the truth might be. However, imaginative conjecture and empirical observation are mutually interdependent episodes. Observations made to test a hypothesis are often the inspiring source of new conjectures or hypotheses. As described by Jacob, the results of an experiment often inspire the modification of a hypothesis and the design of new experiments to test it. The starting point of scientific inquiry is the conception of an idea, a process that is, however, not a subject of investigation for logic or epistemology. The complex conscious and unconscious events underlying the creative mind are properly the interest of empirical psychology. The creative process is not unique to scientists. Philosophers and novelists, poets, and painters are also creative; they, too, advance models of experience and also generalize by induction. What distinguishes science from other forms of knowledge is the process by which this knowledge is justified or corroborated, at least provisionally, by observation and experimentation. THE CRITERION OF DEMARCATION Testing a hypothesis involves at least 4 different activities (Ayala, 1994). First, the hypothesis must be examined for internal consistency. A hypothesis that is self-contradictory or not logically well formed in some other way should be rejected. Second, the logical structure of the hypothesis must be examined to ascertain whether it has explanatory value, i.e., whether it makes the observed phenomena intelligible in some sense, whether it provides an understanding of why the phenomena do in fact occur as observed. A hypothesis that is purely tautological should be rejected because it has no explanatory value. A scientific hypothesis identifies the conditions, processes, or mechanisms that account for the phenomena it purports to explain. Thus, hypotheses establish general relationships between certain conditions and their consequences or between certain causes and their effects. For example, the motions of the planets around the Sun Page 277 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × are explained as a consequence of gravity, and respiration as an effect of red blood cells that carry oxygen from the lungs to various parts of the body. Third, a hypothesis must be examined for its consistency with hypotheses and theories commonly accepted in the particular field of science and to see whether it represents any advance with respect to well-established alternative hypotheses. Lack of consistency with other theories is not always ground for rejection of a hypothesis, although it will often be. Some of the greatest scientific advances occur precisely when it is shown that a widely held and well-supported hypothesis is replaced by a new one that accounts for the same phenomena that were explained by the preexisting hypothesis, and other phenomena it could not account for. One example is the replacement of Newtonian mechanics by the theory of relativity, which rejects the conservation of matter and the simultaneity of events that occur at a distance, 2 fundamental tenets of Newton’s theory. Examples of this kind are pervasive in rapidly advancing disciplines, such as molecular biology at present. The so-called “central dogma” holds that molecular information flows only in one direction, from DNA to RNA to protein. The DNA contains the genetic information that determines what the organism is, but that information has to be expressed in the enzymes (and other proteins) that guide all chemical processes in cells. The information contained in the DNA molecules is conveyed to proteins by means of intermediate molecules, called messenger RNA. David Baltimore (1970) and Howard Temin (Temin and Mizutani, 1970) were awarded the Nobel Prize for discovering independently that information could flow in the opposite direction, from RNA to DNA, by means of the enzyme reverse transcriptase. They showed that some viruses, as they infect cells, are able to copy their RNA into DNA, which then becomes integrated into the DNA of the infected cell, where it is used as if it were the cell’s own DNA. Other examples are the following. Biochemists assumed that only the proteins known as enzymes could catalyze the chemical reactions in cells. However, Thomas Cech (1985) and Sidney Altman received in 1989 the Nobel Prize for independently showing that certain RNA molecules act as enzymes and catalyze their own reactions. One more example concerns the so-called “colinearity” between DNA and protein. Molecular biologists thought that the sequence of nucleotides in the DNA of a gene is expressed consecutively in the sequence of amino acids in the protein. This conception was shaken by the discovery that genes come in pieces, separated by intervening DNA segments that do not code for protein; Richard Roberts and Philip Sharp received the 1993 Nobel Prize for this discovery (Crick, 1979; Chambon, 1981). These revolutionary hypotheses were published after their authors had subjected them to severe empirical tests. Theories that are inconsistent Page 278 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × with well-accepted hypotheses in the relevant discipline are likely to be ignored when they are not availed by convincing empirical evidence. The microhistory of science is littered with farfetched or ad hoc hypotheses, often proposed by individuals with no previous or posterior scientific achievements. Theories of this sort usually fade away because they are ignored by most of the scientific community, although on occasion they engage their interest because the theory may have received attention from the media or even from political or religious bodies. The fiasco 2 decades ago over “cold fusion” was an example of an unlikely and poorly tested hypothesis that received some attention from the scientific community because its proponents were well-established scientists (Taubes, 1993). The fourth and most distinctive step in testing a scientific hypothesis consists of putting the hypothesis on trial by ascertaining whether or not predictions about the world of experience derived as logical consequences from the hypothesis agree with what is actually observed. This is the critical element that distinguishes the empirical sciences from other forms of knowledge: the requirement that scientific hypotheses be empirically falsifiable. Scientific hypotheses cannot be consistent with all possible states of affairs in the empirical world. A hypothesis is scientific only if it is consistent with some but not with other possible states of affairs not yet observed in the world, so that it may be subject to the possibility of falsification by observation. The predictions derived from a scientific hypothesis must be sufficiently precise that they limit the range of possible observations with which they are compatible. If the results of an empirical test agree with the predictions derived from a hypothesis, the hypothesis is said to be provisionally corroborated; otherwise it is falsified. The requirement that a scientific hypothesis be falsifiable has been appropriately called the criterion of demarcation of the empirical sciences because it sets apart the empirical sciences from other forms of knowledge (Popper, 1959, 1963). A hypothesis that is not subject to the possibility of empirical falsification does not belong in the realm of science. The requirement that scientific hypotheses be falsifiable rather than simply verifiable seems surprising at first. It might seem that the goal of science is to establish the “truth” of hypotheses rather than attempt to falsify them, but it is not so. There is an asymmetry between the falsifiability and the verifiability of universal statements that derives from the logical nature of such statements. A universal statement can be shown to be false if it is found to be inconsistent with even 1 singular statement, i.e., a statement about a particular event. But, a universal statement can never be proven true by virtue of the truth of particular statements, no matter how numerous these may be. Consider a particular hypothesis from which a certain consequence is logically derived. Consider now the following argument: If the hypothesis Page 279 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × is true, then the specific consequence must also be true; it is the case that the consequence is true; therefore the hypothesis is true. This is an erroneous kind of inference called by logicians the “fallacy of affirming the consequent.” The error of this kind of inference may be illustrated with the following trivial example: If apples are made of iron, they should fall on the ground when they are cut off a tree; apples fall when they are cut off; therefore, apples are made of iron. The conclusion is invalid even if both premises are true. The reason is that there may be some other explanation or hypothesis from which the same consequences or predictions are derived. The observed phenomena are true because they are consequences from this different hypothesis, rather than from the one used in the deduction. The proper form of logical inference for conditional statements is what logicians call the modus tollens (manner of taking away). It may be represented by the following argument. If a particular hypothesis is true, then a certain consequence must also be true; but evidence shows that the consequence is not true; therefore the hypothesis is false. The modus tollens is a logically conclusive form of inference. If both premises are true, the conclusion falsifying the hypothesis necessarily follows. It follows from this reasoning that it is possible to show the falsity of a universal statement concerning the empirical world; but it is never possible to demonstrate conclusively its truth. This asymmetry between verification and falsification is recognized in the statistical methodology of testing hypotheses. The hypothesis subject to test, the null hypothesis, may be rejected if the observations are inconsistent with it. If the observations are consistent with the predictions derived from the hypothesis, the proper conclusion is that the test has failed to falsify the null hypothesis, not that its truth has been established. Accordingly, scientific theories are never established as definitive truths. As Claude Bernard stated, theories “represent only the current state of our understanding and are bound to be modified by the growth of science” (Bernard, 1865). DARWIN Charles Robert Darwin (1809–1882) was the son and grandson of physicians. In 1825 he enrolled as a medical student at the University of Edinburgh. After 2 years, however, he left Edinburgh to study at the University of Cambridge and prepare to become a clergyman. He was not an exceptional student, but he was deeply interested in natural history. On Dec. 27, 1831, a few months after his graduation from Cambridge, he sailed as a naturalist aboard the HMS Beagle on a round-the-world trip that lasted until October 1836. Darwin was often able to disembark for extended trips ashore to collect natural specimens. The discovery of fossil bones from Page 280 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × large extinct mammals in Argentina and the observation of numerous species of finches in the Galápagos Islands were among the events credited with stimulating Darwin’s interest in how species originate. In 1859 he published On the Origin of Species by Means of Natural Selection (Darwin, 1859), a treatise establishing the theory of evolution and, most important, the role of natural selection in determining its course. He published many other books as well, notably The Descent of Man and Selection in Relation to Sex (Darwin, 1871, 2nd ed.), which extends the theory of natural selection to human evolution. Darwin occupies an exalted place in the history of Western thought, deservedly receiving credit for the theory of evolution. In The Origin of Species, he laid out the evidence demonstrating the evolution of organisms: 2 chapters dedicated to the geological record, 2 chapters dedicated to biogeography, and 1 chapter dedicated to comparative anatomy and embryology (Darwin, 1859, chapters IX–XIII). However, Darwin accomplished something much more important than demonstrating evolution. Indeed, accumulating evidence for common descent with diversification may very well have been a subsidiary objective of Darwin’s masterpiece. Darwin’s Origin of Species is, first and foremost, a sustained argument to solve the problem of how to account scientifically for the “design” of organisms. Darwin seeks to explain the adaptations of organisms, their complexity, diversity, and marvelous contrivances as the result of natural processes. Darwin brings about the evidence for evolution because evolution is a necessary consequence of his theory of natural selection. Nine chapters (I–VIII and XIV) of Origin (Darwin, 1859) are dedicated to natural selection. He explains how natural selection works and the role of hereditary variation (the mechanics of which were not well understood in Darwin’s time), and he considers possible objections to his theory. The evolution of organisms was commonly accepted by naturalists in the middle decades of the 19th century. The distribution of exotic species in South America, in the Galápagos Islands and elsewhere, and the observation of fossil remains of long-extinguished animals during his voyage on the Beagle, would contribute to confirm the reality of evolution in Darwin’s mind. The intellectual challenge after his return to Britain was not simply to accumulate evidence showing that species evolve. Rather, the fundamental challenge was to explain the origin of distinct species of organisms and how they adapted to their environments, that “mystery of mysteries,” as it had been labeled by Darwin’s older contemporary, the prominent scientist and philosopher Sir John Herschel (1792–1871). As Darwin wrote in his Autobiography (Barlow, 1958), “I had always been much struck by such adaptations, and until these could be explained it seemed to me almost useless to endeavor to prove by indirect evidence that species have been modified.” Page 281 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Darwin had come to accept the evolution of organisms by the time he returned from the voyage of the Beagle in October 1836 or shortly thereafter. This is apparent from the notebooks he wrote during the voyage, and those known as the “Transmutation Notebooks,” which he wrote in the ensuing 2 years, after his return (Darwin, 1960). Important as it was to obtain evidence of the origin of species by evolution, this seemed to him to pale compared with the need for demonstrating how the complex adaptations of organisms, their design, came about, namely, by natural selection. The advances of physical science accomplished by the “Copernican Revolution” of the 16th and 17th centuries had brought the workings of the universe under the domain of science: explanation by natural laws that can be tested by observation and experiment. The fundamental commitment was to the postulate that the universe consists of matter in motion governed by natural laws. All physical phenomena could be accounted for as long as the causes became adequately known. However, the origin and configuration of living creatures had been left out, because it seemed that the complex design of organisms could not have come about by chance or by the mechanical laws of physics, chemistry, and astronomy. The notion that the design of organisms could not be accounted for by the laws of nature had been argued at length by philosophers and theologians. William Paley, for example, made the case with considerable biological detail and thoughtful argumentation in his Natural Theology (Paley, 1802), a book that Darwin read as part of his studies at Cambridge University. Paley argued that in the same way that the harmony of the parts making a watch manifest that it had been designed by a skilled watchmaker, so the design of the human eye, with its transparent lens, its retina placed at the precise distance for forming a distinct image, and its large nerve transmitting signals to the brain, manifested to have been designed by the Creator. Darwin’s theory of natural selection brought the adaptations of organisms within the realm of explanation by natural laws. Darwin completed the Copernican Revolution by drawing out for biology the notion of nature as a lawful system of matter in motion that human reason can explain without recourse to supernatural or extranatural agencies. The origin and adaptations of organisms in their profusion and wondrous variations were thus brought into the realm of science. NATURAL SELECTION Darwin considered natural selection, rather than his demonstration of evolution, his most important discovery and designated it as “my theory,” a designation he never used when referring to the evolution of organ- Page 282 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × isms. The discovery of natural selection, Darwin’s awareness that it was a greatly significant discovery because it was science’s answer to Paley’s argument-from-design, and Darwin’s designation of natural selection as “my theory” can be traced in Darwin’s “Red Notebook” and “‘Transmutation Notebooks B to E,” which he started in March 1837, not long after returning (on October 2, 1836) from his 5-year voyage on the Beagle, and completed in late 1839 [Darwin (1960) and see Eldredge (2004)]. Early in the notebooks of 1837–1839, Darwin registers his discovery of natural selection and repeatedly refers to it as “my theory.” From then until his death in 1882, Darwin’s life would be dedicated to substantiating natural selection and its companion postulates, mainly the pervasiveness of hereditary variation and the enormous fertility of organisms, which much surpassed the capacity of available resources. Natural selection became for Darwin “a theory by which to work.” He relentlessly pursued observations and performed experiments to test the theory and resolve presumptive objections. These studies were reported in numerous papers and volumes dedicated to barnacles (fossil and living), orchids and their fertilization by insects, insectivorous and climbing plants, earthworms, and much more. This is how Darwin describes his discovery of natural selection in the Autobiography: “In October, 1838, that is, 15 months after I had begun my systematic enquiry, … it at once struck me that under these circumstances [struggle for existence, as in Malthus] favorable variations would tend to be preserved, and unfavorable ones to be destroyed…. Here then I had at last got a theory by which to work; but I was so anxious to avoid prejudice, that I determined not for some time to write even the briefest sketch of it” (Barlow, 1958). Darwin had in natural selection an explanatory hypothesis to account for the adaptations of organisms that would allow him to design observations and experiments for testing the hypothesis’ validity. “What Darwin meant by ‘a theory by which to work’ was no less than natural selection and trying to derive—as ‘predictions’—the expected consequences of natural selection in action over long periods of time. From natural selection, Darwin tried to derive those very same basic patterns that he had seen in the natural world” (Eldredge, 2004). Despite occasional claims by Darwin himself that he proceeded according to Baconian principles or that he accumulated wholesale facts without any preconceived idea as to what they might imply, Darwin was an excellent practitioner of the hypothetico-deductive method of science. Such claims are little more than “window dressing,” seeking to allay the concerns of his contemporaries, whether philosophers or other possible critics, who would surely find his theory of natural selection hard to take and would be prompt to denounce it as a prejudicial abstraction without Page 283 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × empirical foundation. In his correspondence and Autobiography, Darwin recognized the primary role played by theory. When he came upon the hypothesis of natural selection in 1838, he became aware of its enormous explanatory power to account for the adaptations of organisms and their diversification. He would dedicate much of his scientific activity for the rest of his life to developing the theory of natural selection by considering possible objections and by subjecting it to severe tests, investigating precisely those adaptations (behavioral, sexual, anatomical) that would seem contrived more by preconceived design than as adaptations by natural selection. Modern students of Darwin have convincingly shown Darwin’s exemplary scientific methodology [e.g., De Beer (1964), Mayr (1964, 1982, 1988), Ghiselin (1969), Hull (1973), Ruse (1979), Eldredge (2004)]. Darwin’s 4 monographs on barnacles (Darwin, 1851a,b, 1854a,b) and his books on the fertilization of orchids (Darwin, 1862), human evolution and sexual selection (Darwin, 1871), climbing plants (Darwin, 1875b), insectivorous plants (Darwin, 1875a), the formation of vegetable mold by worms (Darwin, 1881), and others must be seen as severe tests of natural selection, carried out precisely by investigating biological phenomena, including some seemingly quite peculiar, that would seem, at least at first sight, incompatible with his theory of natural selection. Michael Ghiselin (1969) has perceptively shown in The Triumph of the Darwinian Method that the lion’s share of Darwin’s research and publications were a sustained effort to subject the hypothesis of natural selection to severe tests. “Unless one understands this—that Darwin applied, rigorously and consistently, the modern, hypothetico-deductive scientific method—his accomplishments cannot be appreciated. His entire scientific accomplishment must be attributed not to the collection of facts, but to the development of theory…. That Darwin realized the great importance of hypothesis in his work can be documented by his numerous remarks on that subject. In a letter to a colleague, he explicitly compares his hypothesis of natural selection to the undulatory theory of light with its ether, and to the attractive power in Newton’s theory of gravitation” (p. 4). Darwin advanced hypotheses in multiple fields, including geology, plant morphology and physiology, psychology, and evolution, and subjected his hypotheses to empirical test. “The line of argument often pursued throughout my theory is to establish a point as a probability by induction and to apply it as a hypothesis to other parts and see whether it will solve them” [cited in De Beer (1964, p. 94)]. Popper (1959, 1963) has made clear that falsifiability is the criterion of demarcation of the empirical sciences from other forms of knowledge, but also that falsification of seemingly true hypotheses contributes to the advance of science. Darwin recognized the same: “False facts are highly injurious to the progress of Page 284 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × science, for they often endure long; but false views, if supported by some evidence, do little harm, for every one takes a salutary pleasure in proving their falseness; and when this is done, one path toward error is closed and the road to truth is often at the same time opened” (Darwin, 1960). Darwin saw natural selection as an overarching explanatory hypothesis that gave a causal explanation of evolutionary change, was consistent with the experience of plant and animal breeders, and made sense of a host of facts, such as he had uncovered in his research on barnacles, orchids, climbing plants, and many others. The evidence for natural selection, he asserts in a letter is “(i) On its being a vera causa, from the struggle for existence; and the certain geological fact that species do somehow change. (ii) From the analogy of change under domestication by man’s selection. (iii) And chiefly from this view connecting under an intelligible point of view a host of facts” (Darwin F, 1887c, Vol. III, p. 25). PHYLOGENY AND CLASSIFICATION Some philosophers of science have claimed that evolutionary biology is a historical science that does not need to satisfy the requirements of the hypothetico-deductive method. The evolution of organisms, it is argued, is a historical process that depends on unique and unpredictable events, and thus is not subject to the formulation of testable hypotheses and theories. Such claims emanate from a monumental misunderstanding. There are 2 kinds of questions in the study of biological evolution. One concerns history: the study of phylogeny, the unraveling and description of the actual course of evolution on Earth that has led to the present state of the biological world. The scientific disciplines contributing to the study of phylogeny include taxonomy, systematics, paleontology, biogeography, comparative anatomy, comparative embryology, and comparative molecular biology. The second kind of question concerns the elucidation of the mechanisms or processes that bring about evolutionary change. These questions deal with causal, rather than historical, relationships. Population genetics, population ecology, paleobiology, molecular biology, and many other branches of biology are the relevant disciplines. There can be little doubt that the causal study of evolution proceeds by the formulation and empirical testing of hypotheses, according to the hypothetico-deductive methodology that is also characteristic of the physicochemical sciences and other empirical disciplines concerned with causal processes. But the study of evolutionary history is also based on the formulation of empirically testable hypotheses. Consider a simple example. For many years specialists proposed that the evolutionary lineage leading to humans separated from the lineage leading to the great apes (chimpanzee, gorilla, orangutan) before the lineages of the great apes Page 285 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × separated from each other. Some recent authors have suggested instead that humans, chimpanzees, and gorillas are more closely related to each other than the chimpanzee and the gorilla are to the orangutan and other Asian apes. A wealth of empirical predictions can be derived logically from these competing hypotheses. One prediction concerns the degree of similarity between enzymes and other proteins. It is known that the rate of amino acid substitutions is approximately constant when averaged over many proteins and long periods of time. If the older hypothesis is correct, the average amount of protein differentiation should be greater between humans and the African apes than among these and orangutans. However, if the newer hypothesis is correct, humans, gorillas, and chimpanzees should have greater protein similarity than any of the 3 has with orangutans. These alternative predictions provide a critical empirical test of the hypotheses. The available data favor the second hypothesis. Humans, chimpanzees, and gorillas appear to be phylogenetically more closely related to each other than any one of them is related to orangutans, and chimpanzees are more closely related to humans than they are to gorillas. Certain biological disciplines relevant to the study of evolution are largely descriptive and classificatory. Description and classification are necessary activities in all branches of science, but play a greater role in certain biological disciplines, such as systematics and biogeography, than in other disciplines, such as population genetics. Nevertheless, taxonomy, systematics, and biogeography also use the hypothetico-deductive method and formulate empirically testable hypotheses (Brooks and McLennan, 1991; Hillis et al., 1996). Page 286 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × This page intentionally left blank. Page 267 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 268 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 269 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 270 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 271 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 272 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 273 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 274 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 275 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 276 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 277 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 278 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 279 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 280 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 281 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 282 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 283 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 284 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 285 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Page 286 Share Cite Suggested Citation:"13 Darwin and the Scientific Method--Francisco J. Ayala." National Academy of Sciences. 2009. In the Light of Evolution: Volume III: Two Centuries of Darwin. Washington, DC: The National Academies Press. doi: 10.17226/12692. × Next: 14 The Darwinian Revolution: Rethinking Its Meaningand Significance--Michael Ruse » In the Light of Evolution: Volume III: Two Centuries of Darwin Get This Book × Buy Hardback | $80.00 Buy Ebook | $64.99 MyNAP members save 10% online. Login or Register to save! Download Free PDF Two Centuries of Darwin is the outgrowth of an Arthur M. Sackler Colloquium, sponsored by the National Academy of Sciences on January 16-17, 2009. In the chapters of this book, leading evolutionary biologists and science historians reflect on and commemorate the Darwinian Revolution. They canvass modern research approaches and current scientific thought on each of the three main categories of selection (natural, artificial, and sexual) that Darwin addressed during his career. Although Darwin's legacy is associated primarily with the illumination of natural selection in The Origin, he also contemplated and wrote extensively about what we now term artificial selection and sexual selection. In a concluding section of this book, several science historians comment on Darwin's seminal contributions. Two Centuries of Darwin is the third book of the In the Light of Evolution series. Each installment in the series explores evolutionary perspectives on a particular biological topic that is scientifically intriguing but also has special relevance to contemporary societal issues or challenges. The ILE series aims to interpret phenomena in various areas of biology through the lens of evolution and address some of the most intellectually engaging, as well as pragmatically important societal issues of our times. READ FREE ONLINE × ## Welcome to OpenBook! You're looking at OpenBook, NAP.edu's online reading room since 1999. 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How to find the solutions to: [math]\sqrt{x+2}+\sqrt{2-x}=1[/math] - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Survey Question Mathematics Solving Quadratic Equatio... Algebraic Problems Problem Solving Linear Equations Basic Algebra Mathematics Education Arithmetic 5 How do you find the solutions to: √x+2+√2−x=1 x+2+2−x=1? All related (42) Sort Recommended David Smith BSc (Hons) in Mathematics&Computer Science, University of Bristol (Graduated 1986) · Author has 3.6K answers and 4M answer views ·6y Square both sides: 4+2√4−x 2=1 4+2 4−x 2=1 Rearrange: 2√4−x 2=−3 2 4−x 2=−3 Square both sides again: 16−4 x 2=9 16−4 x 2=9 Solve for x:x: x=±√7/4 x=±7/4 Because we squared a couple of times, IT IS VITAL TO CHECK IF OUR SOLUTION WORKS IN THE ORIGINAL PROBLEM! It turns out that neither solution works. But we do find that x=+√7/4 x=+7/4 is a solution of +√x+2−√2−x=1+x+2−2−x=1 x=−√7/4 x=−7/4 is a solution of −√x+2+√2−x=1−x+2+2−x=1 And this happens often when we square away square roots. The potential solutions we find are often not solutions to the original equation, but Continue Reading Square both sides: 4+2√4−x 2=1 4+2 4−x 2=1 Rearrange: 2√4−x 2=−3 2 4−x 2=−3 Square both sides again: 16−4 x 2=9 16−4 x 2=9 Solve for x:x: x=±√7/4 x=±7/4 Because we squared a couple of times, IT IS VITAL TO CHECK IF OUR SOLUTION WORKS IN THE ORIGINAL PROBLEM! It turns out that neither solution works. But we do find that x=+√7/4 x=+7/4 is a solution of +√x+2−√2−x=1+x+2−2−x=1 x=−√7/4 x=−7/4 is a solution of −√x+2+√2−x=1−x+2+2−x=1 And this happens often when we square away square roots. The potential solutions we find are often not solutions to the original equation, but are solutions when we negate one or more of the square roots. In our case the original equation has NO SOLUTIONS, not even in C.C. Upvote · 9 3 Promoted by The Hartford The Hartford We help protect over 1 million small businesses ·Updated Sep 19 What is small business insurance? Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success. The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickl Continue Reading Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. 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Upvote · 999 555 9 1 9 3 Related questions More answers below What is the solution of sqrt (2 + sqrt (2 + sqrt (2 + x))) = x? What are the solutions for √x 2=1 x 2=1? How can I solve √1+√x−2√1−√x=4√1−x 1+x−2 1−x=1−x 4? How do I solve this equation (x+√2)x+(x−√2)x=1(x+2)x+(x−2)x=1? How do I solve this equation x√x+2+x√x−3=2 x+2 x+x−3 x=2? Peter Aouad Director at Cybernet Associates Inc. (2006–present) · Author has 1.8K answers and 1.6M answer views ·6y The given equation has no solution; i.e. f(x) is not equal to zero. f(x) has 2 solutions when “1” is replaced by “2”in your given equation I tweaked your equation to have 2 solutions: x = -2 and x = 2... Continue Reading The given equation has no solution; i.e. f(x) is not equal to zero. f(x) has 2 solutions when “1” is replaced by “2”in your given equation I tweaked your equation to have 2 solutions: x = -2 and x = 2... Upvote · 9 1 Assistant Bot · 1y To solve the equation √x+2+√2−x=1 x+2+2−x=1, we can follow these steps: Isolate one of the square roots: √x+2=1−√2−x x+2=1−2−x 2. Square both sides to eliminate the square root: (√x+2)2=(1−√2−x)2(x+2)2=(1−2−x)2 This simplifies to: x+2=1−2√2−x+(2−x)x+2=1−2 2−x+(2−x) Which further simplifies to: x+2=3−x−2√2−x x+2=3−x−2 2−x 3. Rearranging the equation: 2√2−x=3−x−(x+2)2 2−x=3−x−(x+2) 2√2−x=3−2 x−2 2 2−x=3−2 x−2 2√2−x=1−2 x 2 2−x=1−2 x 4. Divide both sides by 2: √2−x=1−2 x 2 2−x=1−2 x 2 5. Square both sides again: (2−x)=(1−2 x 2)2(2−x)=(1−2 x 2)2 Expanding the ri Continue Reading To solve the equation √x+2+√2−x=1 x+2+2−x=1, we can follow these steps: Isolate one of the square roots: √x+2=1−√2−x x+2=1−2−x 2. Square both sides to eliminate the square root: (√x+2)2=(1−√2−x)2(x+2)2=(1−2−x)2 This simplifies to: x+2=1−2√2−x+(2−x)x+2=1−2 2−x+(2−x) Which further simplifies to: x+2=3−x−2√2−x x+2=3−x−2 2−x 3. Rearranging the equation: 2√2−x=3−x−(x+2)2 2−x=3−x−(x+2) 2√2−x=3−2 x−2 2 2−x=3−2 x−2 2√2−x=1−2 x 2 2−x=1−2 x 4. Divide both sides by 2: √2−x=1−2 x 2 2−x=1−2 x 2 5. Square both sides again: (2−x)=(1−2 x 2)2(2−x)=(1−2 x 2)2 Expanding the right side: 2−x=(1−2 x)2 4 2−x=(1−2 x)2 4 2−x=1−4 x+4 x 2 4 2−x=1−4 x+4 x 2 4 6. Multiply through by 4 to eliminate the fraction: 8−4 x=1−4 x+4 x 2 8−4 x=1−4 x+4 x 2 7. Rearranging gives a quadratic equation: 4 x 2−4 x+1−8+4 x=0 4 x 2−4 x+1−8+4 x=0 4 x 2−7=0 4 x 2−7=0 4 x 2=7 4 x 2=7 x 2=7 4 x 2=7 4 x=±√7 2 x=±7 2 8. Check for extraneous solutions by substituting x=√7 2 x=7 2 and x=−√7 2 x=−7 2 back into the original equation. For x=√7 2 x=7 2: √√7 2+2+√2−√7 2=√√7 2+4 2+√4 2−√7 2 7 2+2+2−7 2=7 2+4 2+4 2−7 2 =√√7+4 2+√4−√7 2=7+4 2+4−7 2 This does not simplify to 1. For x=−√7 2 x=−7 2: √−√7 2+2+√2+√7 2=√2−√7 2+√2+√7 2−7 2+2+2+7 2=2−7 2+2+7 2 This also does not simplify to 1. After checking both potential solutions, you will find that both do not satisfy the original equation, which means the equation has no valid solutions. Thus, the final answer is: There are no solutions to the equation √x+2+√2−x=1 x+2+2−x=1. Upvote · Rory Barrett M.Sc in Mathematics, University of Auckland · Author has 2.1K answers and 1.8M answer views ·6y If you don’t allow negative square roots then there is no solution as sqrt(x+2)>1 if x >=0 and sqrt(2-x) > 1 if x < 0. Look at graph Continue Reading If you don’t allow negative square roots then there is no solution as sqrt(x+2)>1 if x >=0 and sqrt(2-x) > 1 if x < 0. Look at graph Upvote · 9 3 Related questions More answers below How do I solve this equation (√2+√3)x+(√2+√3)x=2 x(2+3)x+(2+3)x=2 x ? What is the solution to |2 x−4|<|2 x−4|<|x−1 x−1|? How would I find a solution for √2+√3+√2+√3+...2+3+2+3+...? Are there any solutions to x 2+x y+y 2=0 x 2+x y+y 2=0? What is x if (√3+√8)x+(√3−√8)x=2(3+8)x+(3−8)x=2? James Gere Upvoted by Krogi Us , M.Sc. Mathematics & Computer Science, Aalto University (1988) · Author has 1.6K answers and 1.3M answer views ·6y Originally Answered: How do I solve this equation (x+\sqrt 2) ^x + (x-\sqrt 2) ^x = 1? · Here is a graph of our “function" expressed as y=(x+√2)x+(x−√2)x−1 y=(x+2)x+(x−2)x−1. The apparent double-valuedness results from the calculator finding both even and odd denominator rational fractional powers of negative numbers in the arguments. Grid-lines fall on the units, and, the extra tick close to the x-axis is my attempt to eye-ball the positive “root" with a floating crosshair. You can read the coordinates below the graph. The truth is: the picture isn't true, but it illustrates a key difficulty that applies to non-integer powers and functions of such that Continue Reading Here is a graph of our “function" expressed as y=(x+√2)x+(x−√2)x−1 y=(x+2)x+(x−2)x−1. The apparent double-valuedness results from the calculator finding both even and odd denominator rational fractional powers of negative numbers in the arguments. Grid-lines fall on the units, and, the extra tick close to the x-axis is my attempt to eye-ball the positive “root" with a floating crosshair. You can read the coordinates below the graph. The truth is: the picture isn't true, but it illustrates a key difficulty that applies to non-integer powers and functions of such that oft leads astray the followers of the Sophmore's Dream. If you look near the origin, the two branches converge on two different limits, corresponding to (√2)0+(−√2)0−1(2)0+(−2)0−1. This is a similar result to using both an even order limit approximation of, 0 0→1 0 0→1, and an odd order approximation of, 0 0→0 0 0→0, to the function, x x x x…x x x x…. If both definitions are tallied in an approximating series, or through iteration, the integral waves back and forth near x = 0 , making convergence impossible, but kind of pretty if you have enough time to run approximations into hundreds of thousands of terms. With our function, the difficulty is setting a proper sign to the powers of negative arguments. To the right of, x=√2 x=2, near a y-value of 4.3509 - 1 , the function gets large quickly but appears continuous and single-valued. We haven't discussed irrational powers. My guess is that this function isn't defined for irrational powers in any compelling way where there are irrational powers of negative arguments giving undecidable signs (Most of the graph above.) So, ultimately, we seem to be looking for a fiction. Oh! yes, I forgot to talk about the complex holes. But, you get the idea. I can't answer exactly what my calculator is graphing because it uses proprietary software. If this was open-source, we might be able to learn exactly what's happening without both of us having to sign non-disclosure contracts. Upvote · 9 6 9 5 9 5 Liam McCann Up to Calculus 3 · Author has 390 answers and 760.1K answer views ·6y Originally Answered: How do you find the solutions to: sqrt(x+2) + sqrt(2-x) = 1? · sqrt(x + 2) + sqrt(2 - x) = 1 (sqrt(x + 2) + sqrt(2 - x))^2 = 1^2 sqrt(x + 2) sqrt(x + 2) + 2 sqrt(x + 2) sqrt(2 - x) + sqrt(2 - x)sqrt(2 - x) = 1 (x+2) + 2 sqrt((x + 2)(2 - x)) + (2 - x) = 1 2 sqrt((x + 2)(2 - x)) + 4 = 1 2 sqrt((x + 2)(2 - x)) = -3 sqrt((x + 2)(2 - x)) = -3/2 We know that a square root cannot result in a negative number, so there are no solutions. We can also confirm this as follows: sqrt(2 - x) does not exist for x > 2. sqrt(x + 2) does not exist for x < -2 Let’s just check our potential integer values in this range: f(x) = sqrt(x + 2) + sqrt(2 - x) x = -2, f(x) = sqrt(0) + Continue Reading sqrt(x + 2) + sqrt(2 - x) = 1 (sqrt(x + 2) + sqrt(2 - x))^2 = 1^2 sqrt(x + 2) sqrt(x + 2) + 2 sqrt(x + 2) sqrt(2 - x) + sqrt(2 - x)sqrt(2 - x) = 1 (x+2) + 2 sqrt((x + 2)(2 - x)) + (2 - x) = 1 2 sqrt((x + 2)(2 - x)) + 4 = 1 2 sqrt((x + 2)(2 - x)) = -3 sqrt((x + 2)(2 - x)) = -3/2 We know that a square root cannot result in a negative number, so there are no solutions. We can also confirm this as follows: sqrt(2 - x) does not exist for x > 2. sqrt(x + 2) does not exist for x < -2 Let’s just check our potential integer values in this range: f(x) = sqrt(x + 2) + sqrt(2 - x) x = -2, f(x) = sqrt(0) + sqrt(4) = 0 + 2 = 2 x = -1, f(x) = sqrt(1) + sqrt(3) = 1 + 1.732 = 2.732 x = 0, f(x) = sqrt(2) + sqrt(2) = 1.414 + 1.414 = 2.828 x = 1, f(x) = sqrt(3) + sqrt(1) = 1.732 + 1 = 2.732 x = 2, f(x) = sqrt(0) + sqrt(4) = 0 + 2 = 2 Nowhere in this range does f(x) ever approach 1. Upvote · 9 4 Promoted by Grammarly Grammarly Great Writing, Simplified ·Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do Continue Reading There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. It’s a writing surface where you can brainstorm, draft, organize your thoughts, and edit—all in one place. It comes with a panel of smart tools to help you refine your work at every step of the writing process and even includes AI Chat to help you get started or unstuck. Expert Review – Your built-in subject expert Need to make sure your ideas land with credibility? Expert Review gives you tailored, discipline-aware feedback grounded in your field—whether you're writing about a specific topic, looking for historical context, or looking for some extra back-up on a point. It’s like having the leading expert on the topic read your paper before you submit it. AI Grader – Your predictive professor preview Curious what your instructor might think? Now, you can get a better idea before you hit send. AI Grader simulates feedback based on your rubric and course context, so you can get a realistic sense of how your paper measures up. It helps you catch weak points and revise with confidence before the official grade rolls in. Citation Finder – Your research sidekick Not sure if you’ve backed up your claims properly? Citation Finder scans your paper and identifies where you need sources—then suggests credible ones to help you tighten your argument. Think fact-checker and librarian rolled into one, working alongside your draft. Reader Reactions – Your clarity compass Writing well is one thing. Writing that resonates with the person reading it is another. Reader Reactions helps you predict how your audience (whether that’s your professor, a TA, recruiter, or classmate) will respond to your writing. With this tool, easily identify what’s clear, what might confuse your reader, and what’s most likely to be remembered. All five tools work together inside Grammarly’s document editor to help you grow your skills and get your writing across the finish line—whether you’re just starting out or fine-tuning your final draft. The best part? It’s built for school, and it’s ready when you are. Try these features and more for free at Grammarly.com and get started today! Upvote · 999 201 99 34 9 3 Sylar Kumar Studied at National Institute of Technology, Patna · Author has 140 answers and 126.1K answer views ·6y Originally Answered: How do you find the solutions to: sqrt(x+2) + sqrt(2-x) = 1? · √2+x+√2−x=1 2+x+2−x=1 (√2+x+√2−x)2=1 2(2+x+2−x)2=1 2 (2+x)+(2−x)+2√(4−x 2)=1(2+x)+(2−x)+2(4−x 2)=1 4+2√4−x 2=1 4+2 4−x 2=1 2√4−x 2=−3 2 4−x 2=−3 4(4−x 2)=9 4(4−x 2)=9 16−4 x 2=9 16−4 x 2=9 7=4 x 2 7=4 x 2 x 2=7/4 x 2=7/4 x=√7/2 x=7/2 or x=−√7/2 x=−7/2 Now we need a sanity check and plug in values of x x into these equations. Without explicit calculation we can see that the first term would be greater than 1, if x x is positive and the second term would be greater than 1 if the x x is negative. Thus, it turns out that both the values of x x are not valid solutions. So x x has no real solutions. Upvote · 9 3 Mohammad Afzaal Butt B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views ·Updated 6y √x+2+√2−x=1 x+2+2−x=1 ⟹√2+x+√2−x=1⟹2+x+2−x=1 squaring both sides.squaring both sides. (2+x)+(2−x)+2√4−x 2=1(2+x)+(2−x)+2 4−x 2=1 ⟹4+2√4−x 2=1⟹4+2 4−x 2=1 ⟹2√4−x 2=−3⟹2 4−x 2=−3 Left hand side is a negative integer.Left hand side is a negative integer. there is no real value of x which can make the left hand side negative.there is no real value of x which can make the left hand side negative. ∴there is no solution to this equation.∴there is no solution to this equation. Upvote · 9 2 9 2 Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) ·Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. Continue Reading This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Upvote · 999 485 999 103 99 17 Sin Keong Tong M.Sc. in Mathematics, University of New South Wales · Author has 283 answers and 361.2K answer views ·6y Originally Answered: How do you find the solutions to: sqrt(x+2) + sqrt(2-x) = 1? · √(x+2)+√(2−x)=1(x+2)+(2−x)=1 Squaring both sides, x+2+2√(x+2)(2−x)+2−x=1 x+2+2(x+2)(2−x)+2−x=1 √(x+2)(2−x)=−3 2(x+2)(2−x)=−3 2 This equation has no solution since √(x+2)(2−x)(x+2)(2−x) is positive by definition. Upvote · Louis M. Rappeport B.S. from University of California, Berkeley · Author has 7.7K answers and 6.7M answer views ·6y √(x+2) + √(2-x)=1 √(x+2)=1-√(2-x) Square both sides: x+2=1–2√(2-x) +2-x 2x-1=-2√(2-x) Square again: 4x²-4x+1=4(2-x) =8–4x 4x²-7=0 4x²=7 x²=7/4 x=±(√7/4) When you plug in these values into the original equation, you get a square root of a negative number. Thus, this equation has no real solutions ……… Upvote · Sponsored by Amazon Business Solutions and supplies to support learning. Save on essentials and reinvest in students and staff. Sign Up 99 85 Paul Grimshaw IT Architect in the Computer Industry (1983–present) · Author has 3.5K answers and 14.9M answer views ·Updated Feb 16 Related How do I solve the equation x 2+√x+1=0 x 2+x+1=0? This obviously has only complex solutions and so to be clear we’re looking to solve: z 2+√z+1=0 z 2+z+1=0 I’ll give an answer and then show how to derive it. If we define: a=√2 3 cos(1 3 arccos(−101 128))−1 3 a=2 3 cos⁡(1 3 arccos⁡(−101 128))−1 3 then there are two complex answers as a conjugate pair: z=−a±i√a 2+1+1 4 a z=−a±i a 2+1+1 4 a These are approximately −0.3438±1.3584 i−0.3438±1.3584 i. One way to get to this and avoid a quartic in complex numbers is to solve the equation as two simultaneous equations in two real variables. Say z=x+y i z=x+y i for real x x and y y, then we can rewri Continue Reading This obviously has only complex solutions and so to be clear we’re looking to solve: z 2+√z+1=0 z 2+z+1=0 I’ll give an answer and then show how to derive it. If we define: a=√2 3 cos(1 3 arccos(−101 128))−1 3 a=2 3 cos⁡(1 3 arccos⁡(−101 128))−1 3 then there are two complex answers as a conjugate pair: z=−a±i√a 2+1+1 4 a z=−a±i a 2+1+1 4 a These are approximately −0.3438±1.3584 i−0.3438±1.3584 i. One way to get to this and avoid a quartic in complex numbers is to solve the equation as two simultaneous equations in two real variables. Say z=x+y i z=x+y i for real x x and y y, then we can rewrite the original equation as the real and imaginary parts expressed in terms of real numbers: R((x+y i)2+√x+y i+1)=0 ℜ((x+y i)2+x+y i+1)=0 I((x+y i)2+√x+y i+1)=0 ℑ((x+y i)2+x+y i+1)=0 To get an idea of what we’re looking at we can graph these two equations: where the blue and red lines are the real and imaginary parts. The (only) solutions in the complex plane are the crossing points, corresponding with the two complex answers I stated above. Moving the root to the rhs and squaring both sides we get ((x+y i)2+1)2−x−y i=0((x+y i)2+1)2−x−y i=0, giving the real and imaginary parts as (x 2−y 2+1)2−4 x 2 y 2−x=0(x 2−y 2+1)2−4 x 2 y 2−x=0 and 4 x y(x 2−y 2+1)−y=0 4 x y(x 2−y 2+1)−y=0. Ignoring the y=0 y=0 possibility in the latter, we can rewrite it as x 2−y 2+1=1/(4 x)x 2−y 2+1=1/(4 x) and also solve this for y 2 y 2. Substituting each of these to eliminate y y in the real part equation we get: 1 16 x 2−4 x 2(x 2−1 4 x+1)−x=0 1 16 x 2−4 x 2(x 2−1 4 x+1)−x=0 This can be rewritten 64 x 6+64 x 4−1=0 64 x 6+64 x 4−1=0, and substituting u−1/3=x 2 u−1/3=x 2 we end up with: u 3−1 3 u+101 12 3=0 u 3−1 3 u+101 12 3=0 Using an approach attributed to François Viète , if we express this equation as: 2 27(4(3 u 2)3−3(3 u 2))+101 12 3=0 2 27(4(3 u 2)3−3(3 u 2))+101 12 3=0 then we can substitute the equality cos(3 arccos(t))=4 t 3−3 t cos⁡(3 arccos⁡(t))=4 t 3−3 t (for −1≤t≤1−1≤t≤1) using t=3 u/2 t=3 u/2 to give us: 2 27(cos(3 arccos(3 u 2))+101 12 3=0 2 27(cos⁡(3 arccos⁡(3 u 2))+101 12 3=0 And we can now solve for u u: u=2 3 cos(1 3 arccos(−101 128))u=2 3 cos⁡(1 3 arccos⁡(−101 128)) (With arccosine being multivalued there are two other solutions, however these are both negative giving imaginary x x values.) Finally translating back to x x we get the a a value above, and we can use the imaginary part equation above to derive the imaginary part y y, noting that we want the negative x x value. Footnotes François Viète - Wikipedia Upvote · 9 3 Max Gretinski Studied Mathematics · Author has 6.6K answers and 2.5M answer views ·Mar 3 Related How can the equation √x+1−√x−1=1 x+1−x−1=1 be solved? In general: Use addition/subtraction to get one radical alone on one side (isolated). Square both sides. This will mean using FOIL on one side, and there will still be one term containing a radical when you are done. Return to step 1. Do steps 1 and 2 again, but this time, there will be no more radicals. Solve the remaining equation. In this case, it will be linear. Check your solution in the original problem, since squaring both sides can cause you to gain false solutions. Let’s try it more generally! √x+a−√x+b x+a−x+b= C Isolate a radical: √x+a x+a= C + √x+b x+b Square: x + a = (C + Continue Reading In general: Use addition/subtraction to get one radical alone on one side (isolated). Square both sides. This will mean using FOIL on one side, and there will still be one term containing a radical when you are done. Return to step 1. Do steps 1 and 2 again, but this time, there will be no more radicals. Solve the remaining equation. In this case, it will be linear. Check your solution in the original problem, since squaring both sides can cause you to gain false solutions. Let’s try it more generally! √x+a−√x+b x+a−x+b= C Isolate a radical: √x+a x+a= C + √x+b x+b Square: x + a = (C + √x+b x+b)(C + √x+b x+b) x + a = C 2 C 2 + 2C√x+b x+b + x + b Move all terms but the radical term to the left side. a - b - C 2 C 2 = 2C√x+b x+b Square both sides. (a−b−C 2)2(a−b−C 2)2 = 4 C 2 4 C 2(x + b) Divide by 4C^2. Then subtract b. x = (a−b−C 2)2 4 C 2(a−b−C 2)2 4 C 2 - b In your case, a = 1, b = -1, and C = 1. This gives us: x = (1+1−1)2 4(1+1−1)2 4 + 1 x = 1 4 1 4 + 1 x = 5 4 5 4 CHECK: √5 4+1−√5 4−1 5 4+1−5 4−1= √9 4−√1 4 9 4−1 4= 3 2−1 2 3 2−1 2= 1 Upvote · 9 4 Bernard Montaron PhD in Mathematics&Discrete Mathematics, Université Pierre Et Marie Curie Paris VI (Graduated 1980) · Author has 3.2K answers and 2.1M answer views ·Feb 6 Related How do I solve the equation x 2+√x+1=0 x 2+x+1=0? This equation does not have real solutions because x 2,√x≥0⟹x 2+√x+1≥1 x 2,x≥0⟹x 2+x+1≥1 Setting x=X 2 x=X 2 this equation becomes X 4+X+1=0 X 4+X+1=0 and it has 4 roots in C C. Using X=±(a+i b)X=±(a+i b) this becomes (a 2−b 2+2 i a b)2±(a+i b)+1=0(a 2−b 2+2 i a b)2±(a+i b)+1=0 Giving the system (a 2−b 2)2−4 a 2 b 2+1±a=0(a 2−b 2)2−4 a 2 b 2+1±a=0 and 4 a b(a 2−b 2)±b=0 4 a b(a 2−b 2)±b=0 We know that b≠0 b≠0 (no real solutions) and the second equation above yields b 2=a 2±1 4 a⟹a 2−b 2=∓1 4 a b 2=a 2±1 4 a⟹a 2−b 2=∓1 4 a Using this in the first equation of the system this yields 1+1 16 a 2−4 a 4=0⟹a 2=0.52872688536918894962...1+1 16 a 2−4 a 4=0⟹a 2=0.52872688536918894962... Finally, we have obtained the four roots x=a x=a Continue Reading This equation does not have real solutions because x 2,√x≥0⟹x 2+√x+1≥1 x 2,x≥0⟹x 2+x+1≥1 Setting x=X 2 x=X 2 this equation becomes X 4+X+1=0 X 4+X+1=0 and it has 4 roots in C C. Using X=±(a+i b)X=±(a+i b) this becomes (a 2−b 2+2 i a b)2±(a+i b)+1=0(a 2−b 2+2 i a b)2±(a+i b)+1=0 Giving the system (a 2−b 2)2−4 a 2 b 2+1±a=0(a 2−b 2)2−4 a 2 b 2+1±a=0 and 4 a b(a 2−b 2)±b=0 4 a b(a 2−b 2)±b=0 We know that b≠0 b≠0 (no real solutions) and the second equation above yields b 2=a 2±1 4 a⟹a 2−b 2=∓1 4 a b 2=a 2±1 4 a⟹a 2−b 2=∓1 4 a Using this in the first equation of the system this yields 1+1 16 a 2−4 a 4=0⟹a 2=0.52872688536918894962...1+1 16 a 2−4 a 4=0⟹a 2=0.52872688536918894962... Finally, we have obtained the four roots x=a 2−b 2+2 i a b x=a 2−b 2+2 i a b x=−0.34381459720147701584...±i 1.35843459972867693702…x=−0.34381459720147701584...±i 1.35843459972867693702… x=0.34381459720147701584...±i 0.62535781178267618013…x=0.34381459720147701584...±i 0.62535781178267618013… Upvote · 9 5 Related questions What is the solution of sqrt (2 + sqrt (2 + sqrt (2 + x))) = x? What are the solutions for √x 2=1 x 2=1? How can I solve √1+√x−2√1−√x=4√1−x 1+x−2 1−x=1−x 4? How do I solve this equation (x+√2)x+(x−√2)x=1(x+2)x+(x−2)x=1? How do I solve this equation x√x+2+x√x−3=2 x+2 x+x−3 x=2? How do I solve this equation (√2+√3)x+(√2+√3)x=2 x(2+3)x+(2+3)x=2 x ? What is the solution to |2 x−4|<|2 x−4|<|x−1 x−1|? How would I find a solution for √2+√3+√2+√3+...2+3+2+3+...? Are there any solutions to x 2+x y+y 2=0 x 2+x y+y 2=0? What is x if (√3+√8)x+(√3−√8)x=2(3+8)x+(3−8)x=2? What are the solutions to 4 x 2−3 x+π=3.14159−3 x+4 x 2 4 x 2−3 x+π=3.14159−3 x+4 x 2? What is the solution for t−1+1000 e−1000 t=0 t−1+1000 e−1000 t=0 ? What are the solutions to s i n(z)=0 s i n(z)=0? How can I find the solution for (y+2 x y 2)d x+(x−2 x 2 y)d y=0(y+2 x y 2)d x+(x−2 x 2 y)d y=0? What are the possible solutions for sin 2 x+cos 2 x=1 sin 2⁡x+cos 2⁡x=1? Related questions What is the solution of sqrt (2 + sqrt (2 + sqrt (2 + x))) = x? What are the solutions for √x 2=1 x 2=1? How can I solve √1+√x−2√1−√x=4√1−x 1+x−2 1−x=1−x 4? How do I solve this equation (x+√2)x+(x−√2)x=1(x+2)x+(x−2)x=1? How do I solve this equation x√x+2+x√x−3=2 x+2 x+x−3 x=2? How do I solve this equation (√2+√3)x+(√2+√3)x=2 x(2+3)x+(2+3)x=2 x ? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://fiveable.me/hs-honors-geometry/unit-3/relationships-lines-planes/study-guide/72e9mdnWxcnJMsLo
printables 🔷Honors Geometry Unit 3 Review 3.1 Relationships between lines and planes 🔷Honors Geometry Unit 3 Review 3.1 Relationships between lines and planes Written by the Fiveable Content Team • Last updated September 2025 Written by the Fiveable Content Team • Last updated September 2025 APA 🔷Honors Geometry Unit & Topic Study Guides 3.1 Relationships between lines and planes 3.2 Angles formed by parallel lines and transversals 3.3 Proving lines parallel or perpendicular 3.4 Equations of parallel and perpendicular lines Lines and planes in 3D space have fascinating relationships. They can be parallel, skew, or intersecting, each with unique properties. Understanding these connections helps visualize complex structures and solve real-world problems. From railroad tracks to helicopter blades, these concepts appear everywhere. Recognizing line-plane orientations is crucial for architecture, engineering, and even everyday tasks like hanging a picture straight on a wall. Lines and Planes in Three-Dimensional Space Types of line-plane relationships Parallel lines Lie in the same plane without intersecting (railroad tracks) Maintain a constant distance between them at all points Skew lines Do not lie in the same plane and never intersect (helicopter blades) Non-parallel and non-intersecting in 3D space Intersecting lines Share a common point of intersection (scissors) Can be coplanar (in the same plane) or non-coplanar (in different planes) Line parallel to a plane Does not intersect the plane at any point (ceiling and floor) Maintains a constant distance from the plane in all directions Line perpendicular to a plane Forms a 90° angle with the plane at the point of intersection (flagpole and ground) Intersects the plane at a single point called the foot of the perpendicular Line intersecting a plane Passes through the plane at a single point (needle and fabric) Forms angles with the plane that are not 90° (acute or obtuse) Classification of line positions Determine the relationship between two lines in 3D space by: Checking if the lines lie in the same plane If they do, they are either parallel or intersecting If they do not, they are skew (like a jungle gym) For lines in the same plane, checking for common points No common points indicate parallel lines A common point indicates intersecting lines (crossroads) Skew lines never have a common point and are not parallel (like telephone wires) Conditions for line-plane orientations A line is parallel to a plane when: The line does not intersect the plane at any point (like a balcony and the ground) The line and plane do not share any common points The distance between the line and plane remains constant (like power lines and the earth) A line is perpendicular to a plane when: The line forms a 90° angle with the plane (like a lamp post and the sidewalk) The line intersects the plane at a single point, the foot of the perpendicular The line is perpendicular to every line in the plane passing through the foot (like a tree trunk and the ground) Applications of line-plane properties Properties of parallel lines and planes: A line parallel to a plane $\implies$ any line $\perp$ to the plane is also $\perp$ to the line Two parallel planes $\implies$ any line $\perp$ to one plane is also $\perp$ to the other A line parallel to a plane $\implies$ any plane containing the line is parallel to the original plane (like sheets of paper in a stack) Properties of perpendicular lines and planes: A line $\perp$ to a plane $\implies$ any line $\perp$ to the line at the intersection point is contained in the plane A line $\perp$ to two intersecting lines in a plane $\implies$ the line is $\perp$ to the plane (like the corner of a room) Two $\perp$ planes $\implies$ any line $\perp$ to one plane at the intersection is contained in the other plane (like the walls and floor of a room)
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https://www.youtube.com/watch?v=I8zz4GFM42k
📚 Proving Midpoint Properties of Trapezoids Study Force 57000 subscribers 5 likes Description 347 views Posted: 10 Jul 2018 ✔ ✔ Ask questions here: ▶ Facebook: ▶ Instagram: ▶ Twitter: Given the coordinates A(−2, −2), B(−1, 2), C(4, 2), D(6,−2), investigate the properties of the midpoints of the non-parallel sides of a trapezoid. The line segment joining the midpoints of the non-parallel sides of a trapezoid is parallel to the parallel sides and has a length equal to the mean of the lengths of the parallel sides. Transcript: in this lesson we'll be proving midpoint properties of trapezoids namely we'll look at how the length of the line segment joining the midpoints of the non parallel sides of a trapezoid is equal to the mean of the lengths of the parallel sides the question reads given the coordinates a b c and d investigate the properties of the midpoints of the non parallel sides of a trapezoid the first thing that I'll do is plot these points our first point is a which is that negative 2 and 2 that's right here our next point is B which is that negative 1 and positive 2 C is at 4 and 2 and D is at 6 and negative 2 let's go ahead and connect these as you can see once we've connected these points this side and this side are parallel what we're investigating here is how the length between the midpoint of this point and this point which is somewhere here connecting a line from here to the midpoint of this side is equal to the average of the lengths of the parallel sides so technically we have to find the midpoint for this side this side find the distance or the length of that line and then find the length of the line for here and here take its average and see how it relates let's go ahead and do that let me start by labeling these points now that I've labeled my trapezoid I'll go ahead and find the midpoint between a and B I'll show you how to do this one and then you can use the same method to find the midpoint for C to D the midpoint is the average of the coordinates so the midpoint between a and B is equal to the average of the X's and the average of the Y's the x-coordinate for a is negative 2 the x-coordinate for B is negative 1/2 then we have negative 2 plus 2 divided by 2 this is equal to 0 negative 2 plus negative 1 is negative 3 over 2 we can plot that and that exists right about here so I'll write that down we have negative 1 point 5 and 0 remember negative 1 point 5 is the same thing as negative 3 over 2 also provided the midpoint between C and D and it happens to have the coordinates 5 and 0 now it's time to find the length from this midpoint to this midpoint to find the lengths we use the distance formula the distance formula is the square root of the difference of the x coordinates so we have x2 minus x1 squared plus the difference of the Y coordinates the x coordinate here is 5 minus negative one point five squared plus y2 is 0 and y one is zero so we don't have to worry about that part now using our calculator we have 5 plus 1.5 to the power of 2 we're going to square root this number we end up with 6.5 so the distance from this midpoint to this midpoint is 6.5 units now we have to find the distance between B and C and the distance between a and D take their average and compare it to 6.5 to find the distance between B and C that's easy we don't need to use that distance formula because we can just use a visual inspection we have 1 2 3 4 5 units and the distance between a and D is 8 units let's add these up 5 plus 8 divide it by 2 because we are taking their average we end up with 13 divided by 2 which is equal to 6.5 therefore to conclude the line segment joining the midpoints of the non parallel sides of a trapezoid is parallel to the parallel sides and has a length equal to the mean of the lengths of the parallel sides so this line which I haven't drawn is parallel to the parallel sides as well
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https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/20%3A_Entropy_and_Free_Energy/20.01%3A_Entropy
Published Time: 2016-06-27T15:54:22Z 20.1: Entropy - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 20: Entropy and Free Energy Introductory Chemistry (CK-12) { } { "20.01:Entropy" : "property get Map 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"authorname:ck12", "source@ "puzzle", "molecular state" ] [ "article:topic", "temperature", "states of matter", "entropy", "kinetic energy", "disorder", "randomness", "showtoc:no", "dissolution", "chemical reactions", "license:ck12", "authorname:ck12", "source@ "puzzle", "molecular state" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Introductory, Conceptual, and GOB Chemistry 4. Introductory Chemistry (CK-12) 5. 20: Entropy and Free Energy 6. 20.1: Entropy Expand/collapse global location Introductory Chemistry (CK-12) Front Matter 1: Introduction to Chemistry 2: Matter and Change 3: Measurements 4: Atomic Structure 5: Electrons in Atoms 6: The Periodic Table 7: Chemical Nomenclature 8: Ionic and Metallic Bonding 9: Covalent Bonding 10: The Mole 11: Chemical Reactions 12: Stoichiometry 13: States of Matter 14: The Behavior of Gases 15: Water 16: Solutions 17: Thermochemistry 18: Kinetics 19: Equilibrium 20: Entropy and Free Energy 21: Acids and Bases 22: Oxidation-Reduction Reactions 23: Electrochemistry 24: Nuclear Chemistry 25: Organic Chemistry 26: Biochemistry Back Matter 20.1: Entropy Last updated Mar 21, 2025 Save as PDF 20: Entropy and Free Energy 20.2: Standard Entropy Page ID 53920 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Entropy 2. Summary When the pieces of a jigsaw puzzle are dumped from the box, the pieces naturally hit the table in a very random pattern. In order to put the puzzle together, a great deal of work must be done to overcome the natural disorder of the pieces. The pieces need to be turned right-side up, then sorted by color or edge (some people like to put the border together first). Finally comes the challenge of finding the exact spot of each piece of the puzzle, in order to obtain the finished picture. Entropy There is a tendency in nature for systems to proceed toward a state of greater disorder or randomness. Entropy is a measure of the degree of randomness or disorder of a system. Entropy is an easy concept to understand when thinking about everyday situations. The entropy of a room that has been recently cleaned and organized is low. As time goes by, it likely will become more disordered and thus its entropy will increase (see figure below). The natural tendency of a system is for its entropy to increase. Figure 20.1.1 20.1.1: The messy room on the right has more entropy than the highly ordered room on the left. Chemical reactions also tend to proceed in such a way as to increase the total entropy of the system. How can you tell if a certain reaction shows an increase or a decrease in entropy? The molecular state of the reactants and products provide certain clues. The general cases below illustrate entropy at the molecular level. For a given substance, the entropy of the liquid state is greater than the entropy of the solid state. Likewise, the entropy of the gas is greater than the entropy of the liquid. Therefore, entropy increases in processes in which solid or liquid reactants form gaseous products. Entropy also increases when solid reactants form liquid products. Entropy increases when a substance is broken up into multiple parts. The process of dissolution increases entropy because the solute particles become separated from one another when a solution is formed. Entropy increases as temperature increases. An increase in temperature means that the particles of the substance have greater kinetic energy. The faster-moving particles have more disorder than particles that are moving slowly at a lower temperature. Entropy generally increases in reactions in which the total number of product molecules is greater than the total number of reactant molecules. An exception to this rule is when a gas is produced from nongaseous reactants. These examples serve to illustrate how the entropy change in a reaction can be predicted: Cl 2(g)→Cl 2(l)Cl 2(g)→Cl 2(l) The entropy is decreasing because a gas is becoming a liquid. CaCO 3(s)→CaO(s)+CO 2(g)CaCO 3(s)→CaO(s)+CO 2(g) The entropy is increasing because a gas is being produced and the number of molecules is increasing. N 2(g)+3 H 2(g)→2 NH 3(g)N 2(g)+3 H 2(g)→2 NH 3(g) The entropy is decreasing because four total reactant molecules are forming two total product molecules. All are gases. AgNO 3(a q)+NaCl(a q)→NaNO 3(a q)+AgCl(s)AgNO 3(a q)+NaCl(a q)→NaNO 3(a q)+AgCl(s) The entropy is decreasing because a solid is formed from aqueous reactants. H 2(g)+Cl 2(g)→2 HCl(g)H 2(g)+Cl 2(g)→2 HCl(g) The entropy change is unknown (but likely not zero), because there are equal numbers of molecules on both sides of the equation, and all are gases. Summary Entropy is defined. Situations involving entropy changes are described. This page titled 20.1: Entropy is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform. LICENSED UNDER Back to top 20: Entropy and Free Energy 20.2: Standard Entropy Was this article helpful? Yes No Recommended articles 20.2: Standard EntropyThis page discusses geothermal energy sources that utilize steam from geysers to fulfill energy needs. It also explains the concept of entropy, which ... 20.3: Spontaneous and Nonspontaneous ReactionsThis page discusses nitroglycerin as an unstable explosive, spontaneous reactions that favor product formation with decreased enthalpy and increased e... 20.4: Free EnergyThis page discusses the steam engine's significance in modern railroads and introduces free energy (G G) in chemical reactions. It explains how free... 20.5: Calculating Free Energy Change (Δ G o)(Δ G o)This page explains the process of baking, emphasizing the importance of heating ingredients to specific temperatures for chemical reactions. It discus... 20.6: Temperature and Free EnergyThis page discusses the reactions of iron ore and coke producing iron and carbon dioxide at high temperatures. It also explains the decomposition of c... Article typeSection or PageAuthorCK-12 FoundationLicenseCK-12Show Page TOCno on page Tags chemical reactions disorder dissolution entropy kinetic energy molecular state puzzle randomness source@ states of matter temperature © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 20: Entropy and Free Energy 20.2: Standard Entropy
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https://www.wyzant.com/resources/answers/947212/integration-with-partial-fractions
WYZANT TUTORING Rozhan F. Integration with partial fractions Make a substitution to express the integrand as a rational function and then evaluate the integral. (Remember the constant of integration.) ∫ dx/(x√x-1) 3 Answers By Expert Tutors Doug C. answered • 10/08/24 Math Tutor with Reputation to make difficult concepts understandable Let u = √(x-1) u2 = x-1 x = u2 + 1 dx = 2udu ∫ (2udu)/[((u2+1)u] = 2∫du/(u2+1) = 2tan-1(u) = 2tan-1(√(x-1))+C desmos.com/calculator/nd9ilu6w19 Paul M. answered • 10/08/24 B.S. in Mathematics & M.D. I would make the substitution y2=x-1 so that x=y2+1 and dx=2y2 dy Then the integration can be performed by a simple u-substitution. Yefim S. answered • 10/08/24 Math Tutor with Experience x = t2; dx = 2tdt; I = ∫dx/(x√x - 1) = ∫2tdt/(t3 - 1) = ∫2tdt/(t - 1)(t2 + t + 1); 2t/(t - 1)(t2 + t + 1) = a/(t - 1) + (bt + c)/(t2 + t + 1); 2t = a(t2 + t + 1) + (bt + c)(t - 1); a + b = 0; a - b + c = 2; a - c = 0; c = a; b = - a; 3a = 2; a = 2/3; b = - 2/3, c = 2/3 I = ∫(2/3)dt/(t - 1) - 2/3∫(t - 1)dt/(t2 + t + 1) = 2/3lnIt - 1I - 1/3∫(2t +1 - 3)dt/(t2 + t + 1) = 2/3lnIt - 1I - 1/3ln(t2 + t + 1) + ∫dt/[(t + 1/2)2 + 3/4] = 2/3lnI√x -1I - 1/3ln(x + √x + 1) + 2/√3tan-1[( √x +1 /2)2/√3] + C Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. RELATED TOPICS RELATED QUESTIONS CAN I SUBMIT A MATH EQUATION I'M HAVING PROBLEMS WITH? Answers · 3 If i have rational function and it has a numerator that can be factored and the denominator is already factored out would I simplify by factoring the numerator? Answers · 7 how do i find where a function is discontinuous if the bottom part of the function has been factored out? Answers · 3 find the limit as it approaches -3 in the equation (6x+9)/x^4+6x^3+9x^2 Answers · 8 prove addition form for coshx Answers · 4 RECOMMENDED TUTORS Harry O. David W. Whitney T. find an online tutor Download our free app A link to the app was sent to your phone. Get to know us Learn with us Work with us Download our free app Let’s keep in touch Need more help? Learn more about how it works Tutors by Subject Tutors by Location IXL Comprehensive K-12 personalized learning Rosetta Stone Immersive learning for 25 languages Education.com 35,000 worksheets, games, and lesson plans TPT Marketplace for millions of educator-created resources Vocabulary.com Adaptive learning for English vocabulary ABCya Fun educational games for kids SpanishDictionary.com Spanish-English dictionary, translator, and learning Inglés.com Diccionario inglés-español, traductor y sitio de aprendizaje Emmersion Fast and accurate language certification
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https://www.sciencedirect.com/science/article/pii/S2211568416302339
Skip to article My account Sign in View PDF Diagnostic and Interventional Imaging Volume 97, Issue 12, December 2016, Pages 1241-1257 Continuing education program: focus€¦ Imaging of neuroendocrine tumors of the pancreas Author links open overlay panel, , , , , , rights and content Under an Elsevier user license Open archive Abstract Pancreatic neuroendocrine tumors (PNETs) are rare and represent a heterogeneous disease. PNET can be functioning or non-functioning with different clinical presentations and different prognosis based on WHO and pTNM classifications. The role of imaging includes the localization of small functioning tumor, differentiation of these tumors from adenocarcinoma, identification of signs of malignancy and evaluation of extent. PNETs have a broad spectrum of appearance. On CT and MRI, most of functioning PNETs are well defined small tumors with intense and homogeneous enhancement on arterial and portal phases. However, some PNETs with a more fibrous content may have a more delayed enhancement that is best depicted on the delayed phase. Other PNETs can present as purely cystic, complex cystic and solid tumors and calcified tumors. Non-functioning PNETs are larger with less intense and more heterogeneous enhancement. Functional imaging is useful for disease staging, to detect disease recurrence or the primary but also to select patient candidate for peptide receptor radiometabolic treatment. Somatostatin receptor scintigraphy (SRS) (Octreoscan®) is still the most available technique. Gallium 68-SST analogue PET have been demonstrated to be more sensitive than SRS-SPEC and it will be the future of functional imaging for NET. Finally, 18FDG PET/CT is indicated for more aggressive PNET as defined either by negative SRS and huge tumor burden or ki67 above 10% or poorly differentiated PNEC tumors. Keywords Neuroendocrine tumor Pancreas Pancreatic malignancies Computed tomography Magnetic resonance imaging Cited by (0) © 2016 Editions françaises de radiologie. Published by Elsevier Masson SAS. All rights reserved.
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https://www.frontiersin.org/journals/genetics/articles/10.3389/fgene.2022.1013858/full
Frontiers | Using dried blood spot on HemoTypeSC™, a new frontier for newborn screening for sickle cell disease in Nigeria Frontiers in Genetics About us About us Who we are Mission and values History Leadership Awards Impact and progress Frontiers' impact Our annual reports Publishing model How we publish Open access Peer review Research integrity Research Topics FAIR² Data Management Fee policy Services Societies National consortia Institutional partnerships Collaborators More from Frontiers Frontiers Forum Frontiers Planet Prize Press office Sustainability Career opportunities Contact us All journalsAll articlesSubmit your researchSearchLogin Frontiers in Genetics Sections Sections Applied Genetic Epidemiology Behavioral and Psychiatric Genetics Cancer Genetics and Oncogenomics Computational Genomics Cytogenomics ELSI in Science and Genetics Epigenomics and Epigenetics Evolutionary and Population Genetics Genetics of Aging Genetics of Common and Rare Diseases Genomic Assay Technology Genomics of Plants and the Phytoecosystem Human and Medical Genomics Immunogenetics Livestock Genomics Neurogenomics Nutritional Genomics Pharmacogenetics and Pharmacogenomics RNA Statistical Genetics and Methodology Stem Cell Research Toxicogenomics ArticlesResearch TopicsEditorial board About journal About journal Scope Field chief editors Mission & scope Facts Journal sections Open access statement Copyright statement Quality For authors Why submit? Article types Author guidelines Editor guidelines Publishing fees Submission checklist Contact editorial office About us About us Who we are Mission and values History Leadership Awards Impact and progress Frontiers' impact Our annual reports Publishing model How we publish Open access Peer review Research integrity Research Topics FAIR² Data Management Fee policy Services Societies National consortia Institutional partnerships Collaborators More from Frontiers Frontiers Forum Frontiers Planet Prize Press office Sustainability Career opportunities Contact us All journalsAll articlesSubmit your research Frontiers in Genetics Sections Sections Applied Genetic Epidemiology Behavioral and Psychiatric Genetics Cancer Genetics and Oncogenomics Computational Genomics Cytogenomics ELSI in Science and Genetics Epigenomics and Epigenetics Evolutionary and Population Genetics Genetics of Aging Genetics of Common and Rare Diseases Genomic Assay Technology Genomics of Plants and the Phytoecosystem Human and Medical Genomics Immunogenetics Livestock Genomics Neurogenomics Nutritional Genomics Pharmacogenetics and Pharmacogenomics RNA Statistical Genetics and Methodology Stem Cell Research Toxicogenomics ArticlesResearch TopicsEditorial board About journal About journal Scope Field chief editors Mission & scope Facts Journal sections Open access statement Copyright statement Quality For authors Why submit? Article types Author guidelines Editor guidelines Publishing fees Submission checklist Contact editorial office Frontiers in Genetics Sections Sections Applied Genetic Epidemiology Behavioral and Psychiatric Genetics Cancer Genetics and Oncogenomics Computational Genomics Cytogenomics ELSI in Science and Genetics Epigenomics and Epigenetics Evolutionary and Population Genetics Genetics of Aging Genetics of Common and Rare Diseases Genomic Assay Technology Genomics of Plants and the Phytoecosystem Human and Medical Genomics Immunogenetics Livestock Genomics Neurogenomics Nutritional Genomics Pharmacogenetics and Pharmacogenomics RNA Statistical Genetics and Methodology Stem Cell Research Toxicogenomics ArticlesResearch TopicsEditorial board About journal About journal Scope Field chief editors Mission & scope Facts Journal sections Open access statement Copyright statement Quality For authors Why submit? Article types Author guidelines Editor guidelines Publishing fees Submission checklist Contact editorial office Submit your researchSearchLogin Your new experience awaits. Try the new design now and help us make it even better Switch to the new experience ORIGINAL RESEARCH article Front. Genet., 26 October 2022 Sec. Genetics of Common and Rare Diseases Volume 13 - 2022 | This article is part of the Research Topic Building Capacity for Sickle Cell Disease Research and HealthcareView all 16 articles Using dried blood spot on HemoTypeSC™, a new frontier for newborn screening for sickle cell disease in Nigeria Chinwe O. Okeke1,2Reuben I. Chianumba1Hezekiah Isa1,3Samuel Asala1,4Obiageli E. Nnodu1,3 1 Center of Excellence for Sickle Cell Research and Training, University of Abuja, Abuja, Nigeria 2 Department of Medical Laboratory Science, Faculty of Health Sciences, University of Nigeria Nsukka, Nsukka, Enugu, Nigeria 3 Department of Haematology and Blood Transfusion, College of Health Sciences, University of Abuja, Abuja, Nigeria 4 Department of Anatomical Sciences, College of Health Sciences, University of Abuja, Abuja, Nigeria Background: HemoTypeSC is a rapid, point-of-care testing (POCT) device for sickle cell disease (SCD) that traditionally uses the capillary blood from heel stick collected at the point of testing, a procedure that makes mass screening cumbersome and less cost-effective. Using dried blood spots (DBS) on HemoTypeSC could mitigate this challenge. Therefore, this study aimed to determine the feasibility of eluting blood from DBS to read on HemoTypeSC. Methods: DBS and fresh samples from heel sticks were collected from 511 newborns at the immunization clinics of six Primary Health Centers in Abuja, Nigeria. The two samples from each newborn were analyzed using HemoType SC and then compared with the result of the isoelectric focusing (IEF) test. Results: Of the 511 newborns, 241 were males and 270 were females. Standard HemoTypeSC (using fresh samples collected from heel sticks) and HemoTypeSC using DBS identified 404 (79.0%) HbAA, 100 (19.6%) HbAS, 6 (1.2%) HbSS, and 1 (0.2%) HbAC phenotypes. The IEF tests identified 370 (72.4%) HbAA, 133 (26.0%) HbAS, 5 (1.0%) HbSS, and 3 (0.6%) HbAC phenotypes. The sensitivity, specificity, positive predictive value (PPV), negative predictive value (NPV) and overall accuracy of HemoTypeSC using DBS, compared to standard HemoTypeSC POCT was 100%. IEF method showed for AA, AS, AC phenotypes; sensitivity; 84.7%, 67%,100% respectively, specificity; 67.6%, 86%, 99% respectively, PPV; 91.2%, 53%, 50% respectively, NPV; 52.7%, 91%, 100% respectively. For SS phenotype, IEF showed 100% specificity, sensitivity, PPV and NPV. Conclusion: HemoTypeSC test using dried blood spot is as accurate as the standard point-of-care HemoTypeSC test. The use of DBS on HemoTypeSC could ensure better efficiency and cost-effectiveness in mass newborn screening for SCD. 1 Introduction Sickle cell disease (SCD) is a genetic blood disorder with high prevalence in Sub-Saharan Africa (Nnodu et al., 2021). It is estimated that 3,12,000 newborns were born with sickle cell anemia globally in 2010, with 2,30,000 being born in Sub-Saharan Africa, accounting for 80 percent of the global sickle cell anaemia population (Berger et al., 2022) (Lanzkron, Patrick Carroll and Haywood, 2013). In high-income countries, the life expectancy of SCD patients has increased dramatically over the last 40 years, reaching 50 years. Whereas in Sub-Saharan Africa, most children with SCD are thought to die before reaching the age of five (Ware et al., 2017). Predictably between 2010 and 2050, the overall number of births affected by SCD will be 14,242,000. Specifically for Nigeria, the number is likely going to rise from 91,000 newborns with SCA in 2010 to 1,40,800 with SCA in 2050 (Piel et al., 2013). It is expected that large-scale universal screening stands the chance of saving up to 9,806,000 newborns with SCA globally, 85% of these newborns will be born in sub-Saharan Africa (Piel et al., 2013). SCD burden is high in Africa with especially high mortality amongst the under-fives. The prevalence of sickle cell trait in Nigeria is 25% and that of homozygous state is up to 2% in some regions. Nigeria is the country that has the highest burden of SCD (NDHS, 2018). Model estimates from the Nigeria National Demographic Survey showed that the national average under-5 mortality for children with SCD born between 2003 and 2013 was 490 per 1,000 livebirths (95% CI 270–700), 4·0 times higher (95% CI 2·1–6·0) than children with HbAA, with about 4·2% (95% CI 1·7–6·9) of national under-5 mortality attributable to excess mortality from SCD (Hsu et al., 2018) (Nnodu et al., 2021). In high-prevalence areas, there is evidence of several benefits of universal newborn screening (NBS) for SCD (Green et al., 2016). Except for Egypt, many African nations lack a national NBS program. In the past, the Republics of Benin and Ghana were the only countries in Africa with SCD NBS programs, and even those are not at national levels despite the burden of SCD on the continent and the benefits of NBS with SCD management (Rahimy et al., 2009). Activities in the other countries include a variety of NBS pilot studies (Therrell et al., 2020). Thus, there is a need for NBS programs to be scaled up nationally in most African countries. In Nigeria, the groundwork for a nationwide program has already been laid, but hampered by inadequate funding, high cost of reagents, and a lack of skilled manpower amongst other obstacles (Hsu et al., 2018). The fact that the bulk of the people in SSA live in rural regions and lacks access to healthcare is one of the most significant difficulties (National Population Commission (NPC) [Nigeria] and ICF, 2019). Point-of- care test (POCT) devices are reliable, easy-to-use and cheap, hence can considerably facilitate the identification of individuals with SCD in Nigeria and other countries in which the SCD prevalence is high (Nnodu et al., 2019). A few POCT devices for SCD have recently been developed based on differential erythrocyte density (Kumar et al., 2014), differential mobility of Hb S and Hb A through filter paper (Yang et al., 2013), and a polyclonal antibody-based capture immunoassay (Kanter et al., 2015). All of these have one challenge or the other. Some of these challenges are either that the devices require apparatus as an inherent element of the technique to attain maximal specificity and sensitivity, or because of their lack of accuracy (Bond et al., 2017). A unique POCT (HemoTypeSC™ uses monoclonal antibodies (MAb) to distinguish between normal adult haemoglobin (HbA), sickle haemoglobin (HbS), and haemoglobin C (HbC) (Quinn et al., 2016). One of the first reports was in the evaluation of 100 whole blood samples from individuals with common relevant Hb phenotypes. HemoTypeSC was proven to be 100 percent accurate in identifying the proper Hb phenotype (Quinn et al., 2016). Since these antibodies are blind to haemoglobin F (HbF), they can reliably diagnose neonates with increased HbF but low levels of HbA or HbS. In a study by Nnodu et al. (2019), the overall accuracy, specificity, and sensitivity of HemoTypeSC in identifying Hb phenotypes (AA, AS, AC, SS, SC, and CC) across multiple Nigerian primary healthcare centers in a real-life, field setting were evaluated. The results obtained in this study corroborated previously published findings and revealed a sensitivity and specificity of 100 percent for HbS and HbC, using high-performance liquid chromatography (HPLC) method as gold standard. Dried blood spot (DBS) is a minimally invasive blood sampling technique. Blood samples are collected from the heel of newborns and applied into a cellulose or polymer card paper. The blood loaded card paper is air dried, after which it is stored in low gas-permeability plastic bags containing desiccant to reduce humidity. DBS is one of the most convenient tools for blood sample collection. Its benefits include; analytical measurements for more than 50 (Fifty) analytes, the sample has been found to be stable for a couple of months at ambient temperature or refrigeration with loss of enzymatic activity to a negligible extent, easy shipment zip-lock bags requiring no cold chain from sampling point to the laboratory and reduced risk of infection as a result of contaminated samples (Saud, 2018). Thus, this is an ideal sampling method in resource-poor settings. DBS sample has an economical preference for many clinical applications (Chace and Hannon, 2016). DBS has been used successfully on isoelectric focusing method (IEF) (Williams, 2016), and HPLC (Inusa et al., 2015). HemoTypeSC is one of the POCT devices for SCD that has been extensively investigated and found with commendable performance characteristics. The normal HemoTypeSC procedure makes use of fresh capillary blood; hence screening has to be on the spot. Moreover, to reduce the turnaround time, two or three personnel have to be involved in the process, hence making it more cost implicative. These factors reduce the general effectiveness of the normal HemoTypeSC™ technique for use in a mass screening settings like immunization centers in resource limited countries. Using fresh capillary blood sample for running HemoTypeSC technique may not provide the required efficiency needed in a mass screening setting. Considering the afore stated challenges, Dry blood sampling may be the way forward. Here, we tried to determine the possibility of eluting blood from DBS to read HemoType SC™ compared to the standard method of using fresh capillary blood as applied in POCT. 2 Materials and methods This is a pilot study and the aim is to find out the possibility of eluting blood from DBS to run HemoTypeSC™ protocol and to compare the results obtained with standard HemoTypeSC™ POCT and IEF method. Newborns zero (0) to six (6)weeks of age drawn across six immunization centers in the Federal capital territory (FCT) Abuja participated in the study. 2.1 The test methods 2.1.1 HemoTypeSC™ Monoclonal antibodies (Mab) are used in the competitive lateral flow immunoassay known as HemoTypeSC™ to detect the presence of hemoglobin A, S, and C. The hemoglobin phenotypes HbAA, HbSS, HbSC, HbCC, HbAS, and HbAC are quickly detected using it (Bassimbié Kakou Danho et al., 2021). Each MAb bound just its target in a competitive enzyme-linked immunosorbent test with just 1.0% cross-reactivity. Since these antibodies are blind to haemoglobin F (HbF), it is possible to diagnose neonates with elevated HbF and low levels of HbA or HbS (Nnodu et al., 2019). A cellulose wick, antibody-impregnated nitrocellulose, and laminated fiberglass sample pads make up test strips, which allow liquid samples to pass through the three components in a particular order. The process involved rehydrating the dried gold conjugate and dilution of the lysed blood sample using an assay solution that contained detergents and non-specific blocking reagents (Quinn et al., 2016). The presence of a line on the strips indicates the absence of the hemoglobin variant in the blood sample (Bassimbié Kakou Danho et al., 2021). 2.1.2 Isoelectric focusing IEF employs an agarose gel that enables qualitative and semi-quantitative analysis by separating various haemoglobins from a patient sample into distinct bands based on their isoelectric point. Haemoglobins are separated on one axis using IEF gel. Visual comparison of the individual bands to the closest reference samples is a typical method of qualitatively measuring patient sample. 2.1.3 High-performance liquid chromatography The principle of HPLC is based on the distribution of the analytes between a stationary phase such as the packing in a column and a mobile phase which is the sample or analytes which is pumped through a valve at high pressure. The interaction between the sample and the stationary phase or column depends on the chemical structure of the analyte which allow some molecules to be retained while some pass through more easily. The analyte is detected after leaving the column with the signals converted and recorded by a computer software in the form of a graph in wavelengths called a chromatograph. This method can be used to separate and quantitate haemoglobin and its variants. It is particularly sensitive to the detection Hb A2, Hb F. Ethical clearance was obtained from Federal Capital Territory Research Ethics Committee. 511 newborns were tested at 6 immunization centers in the FCT. The sampling was carried out between October 2021 and January 2022. Mothers of all eligible babies coming for immunization at the centers were approached for testing. Informed signed consent was obtained. The “Standard Precautions” protocol developed by the US Centers for Disease Control and Prevention was followed throughout the sample collection and testing to prevent infection when working with human blood samples (Quinn et al., 2016). 2.2 Storage, sampling and testing The HemoTypeSC test kits containing the lateral flow assay (LFA) test strip, a transfer pipette, a sample cup, and a volumetric inoculation loop were stored at room temperature. HemoTypeSC is considered to be stable in high heat and does not require refrigeration. Blood samples from babies 6 weeks and below were drawn by heel-prick into labeled filter paper cards unto a HemoTypeSC™ POCT Sample collection strip, supplied by Silver Lake Research Corporation. The POCT was carried out on site, while the blood spots were air dried for a minimum of 3 h at 18°C–25°C, shipped to the Centre of Excellence for Sickle Cell Disease Research and Training (CESRTA), University of Abuja Newborn Screening Laboratory and stored in gas-impermeable zipper bag, containing desiccant sachets and kept in the Refrigerator at −20°C. Iso electric focusing testing was performed at CESRTA lab using DBS. After 1 week, the dried blood sample was eluted and HemoTypeSC™ standard protocol followed to determine the test result. The tests were carried out strictly following the manufacturer’s instructions and test results were interpreted based on a reference chart provided by the manufacturer. Clinical control samples of previously-diagnosed AA, AS, SS, and SC individuals were included with each batch of HemoTypeSC and IEF to assess the performance of these techniques. Results from HemoTypeSC standard POCT, HemoTypeSC using DBS and IEF were then compiled in a spreadsheet for analysis. 2.3 Assessment The sensitivity, specificity, positive and negative predictive values, and overall accuracy of HemoTypeSC using DBS was compared to standard HemoTypeSC POCT and IEF were calculated. Sensitivity was defined as 100% × TP/(FN + TP) specificity as 100% × TN/(FP + TN), positive predictive value as 100% × TP/(TP + FP), negative predictive value as 100% × TN/(TN + FN) and overall accuracy as (prevalence × sensitivity)/(1—prevalence) (specificity), where TP = number of true positive events, FP = number of false positive events, and TN = number of true negative events . 3 Results A number of 241 males and 270 females were screened. The HemoTypeSC standard POCT protocol tests identified 404 HbAA (79.0%), 100 HbAS (19.6%), 6 HbSS (1.2%), and 1 HbAC (0.19%). HemoTypeSC using DBS showed the same result pattern as that done using the standard POCT protocol. The test cannot differentiate Hb SS and sickle -β 0- thalassemia. No HbCC or HbSC were identified. Details per center and allele frequencies are presented in Table 1. The IEF tests identified 370 HbAA (72.4%), 133 HbAS (26.0%), 5 HbSS (0.97%), and 3 HbAC (0.58%). The results of the 84 discordant samples are displayed on Tables 2, 3 gives a summary of the frequency of the various haemoglobin phenotypes. Specificity, positive predictive value (PPV), negative predictive value (NPV) and overall accuracy of HemoTypeSC using dried blood spot, compared to standard HemoTypeSC were 100% as seen in Table 4. Isoelectric focusing (IEF) method showed; for AA, AS, AC Sensitivity; 84.7, 67, 100 respectively specificity; 67.6, 86, 99 respectively. Positive predictive value; 91.2, 53, 50 respectively. Negative predictive value; 52.7, 91, 100. For SS phenotype, IEF showed 100% specificity, sensitivity, positive predictive value and negative predictive value. Discordant results were found for a total of 84 samples. These 84 discordant samples were run with HPLC and the following results were obtained: 57 (AA), 21 (AS),1 (SS),1 (A3), 2 (ACS), 2 (DA). The discordant results were analysed by HPLC which showed that HemoTypeSC correctly identified all the HbAA, 21 of the HbAS but categorized 3 AD, 1 AC5, and 1 A3 as AS while IEF failed to identify 30 HbAA, wrongly labelling them as AS and was the only method to report 2 AC. Table 5 shows the measurement of agreement of Kappa between standard HemoTypeSC POCT and DBS HemoTypeSC and the measurement of agreement of Kappa between DBS HemoTypeSC and IEF. The table revealed Kappa value between HemoTypeSC POCT and DBS HemoTypeSC as 1.000 showing a strong agreement. The Kappa value between DBS HemoTypeSC and IEF was 0.540 showing a moderate agreement. For both measurements p-value <0.05 as seen in Table 5. TABLE 1 TABLE 1. Genotype and allele frequencies identified by HemoTypeSC and IEF in the 6 centers comprising University of Abuja Teaching Hospital immunization centers attached to it. TABLE 2 TABLE 2. Showing the results of the 84 discordant samples run with HPLC. TABLE 3 TABLE 3. Showing the frequency distribution of the Hb phenotypes studied in the three methods used. TABLE 4 TABLE 4. Performance characteristics of DBS/POCT HemoTypeSC compared to IEF. TABLE 5 TABLE 5. Cohen’s Kappa Statistics showing the level of agreement of Kappa between DBS HemoTypeSC/Standard HemoTypeSC and DBS HemoTypeSC/IEF. Results revealed that most of the HPLC results of the discordant samples agree with the standard POCT and DBS HemoTypeSC results and not with the IEF results thus calling into question the validity of the designation of IEF/HPLC as gold standard methods (Nnodu et al., 2019, Nnodu et al., 2020). 4 Discussion Our findings agree with previous studies suggesting a specificity and specificity of 100% (Nnodu et al., 2019) (Olatunya et al., 2021) for HbS and HbC in ideal conditions using the standard HemoTypeSC POCT procedure. Nnodu et al., worked on 1,121 samples and compared HemoTypeSC POCT with HPLC (GOLD Standard). They got a sensitivity of; 0.989 (AA), 0.983 (AS), 1.000 (SS), 0.933 (AC), and Specificity of: 0.993 (AA), 0.992 (AS), 0.999 (SS), 1.000 (AC). Olatunya et al. (2021) in 2021 compared Cellulose Acetate Electrophoresis, HemoTypeSC POCT and HPLC methods with PCR in 158 participants. HemoTypeSC showed both a Sensitivity and Specificity of 1.00 (100%) for AA, AC, AS, SC, and SS. Using DBS to run HemoTypeSC, our study produced a result that is 100% concordant with the standard HemoTypeSC procedure. This means that DBS samples can conveniently run on HemoTypeSC and accurate results produced. SSA countries like Nigeria, with high burden of SCD are faced with challenges of lack of accessibility to healthcare facilities for large portions of their populations. In such situations, primary care must become the focal point of SCD screening and treatment, with focus on initiatives that utilize user-friendly, reasonably priced technology and engage a sizable percentage of the community). No doubt certain attributes and qualities of POCT devices for sickle cell disease screening make it very suitable for use in SSA. These qualities are; kit not requiring rigorous procedure and expensive reagents, special skills and electricity not required. The 84 Discordant results were run with HPLC which showed that HemoTypeSC correctly identified all the HbAA, 21 of the HbAS but categorised 3 AD, 1 AC5, and 1 A3 as AS. The results obtained by HPLC tally more with HemoTypeSC than with IEF as can be seen from Table 2; HPLC recorded 57 AA and HemoTpyeSC recorded 57 by both DBS and POCT methods, IEF recorded 27. HPLC recorded 21 AS with 5 variants; HemoTypeSC recorded 26 AS while IEF had 55 AS. From the results by the three methods, the ones by IEF seem as the outlier. Obviously using HPLC as gold standard, and referring to Table 3, IEF mis-identified 33 babies who had other phenotypes as having sickle cell trait. There are still gaps in effective application of POCT in real life mass screening settings like immunization centers. As new knowledge is explored and meaningful collaborations developed, a situation arises when a battery of tests needs to be conducted as is applicable in the advanced countries, the use of these devices as POCT will not suffice, hence the application of DBS on such devices as HemoTypeSC will bridge these gaps. 4.1 Using dried blood spots on HemoTypeSC Using DBS on HemoTypeSC has the potential of making the move of scaling up efforts to adopt early diagnosis, penicillin prophylaxis and hydroxyurea therapy, to forestall under5 mortality in SSA a reality. High level of discordance was discovered when DBS HemoTypeSC™ and IEF were compared on same subjects (16.4%). Among the 84 discordant samples, the AA and AS phenotype results by HPLC were in agreement with the standard HemoTypeSC POCT result. Only one positive AS sampled agreed with IEF. This goes to reveal the challenge associated with the high technicality involved with IEF and we infer that the IEF run in the Low- and middle-income countries may not be as efficient as those in the developed or high-income countries IEF method is the gold standard for newborn screening, the inconsistency observed in this study between IEF results and HPLC might be due to technical challenges facing the use of IEF in resource poor settings. Since DBS is utilized in other public health initiatives like HIV screening (Sikombe Id et al., 2019), it is possible that samples from early newborn diagnosis might be quickly and affordably tested for sickle hemoglobin using this technique. Sickle SCAN which is also is a rapid, qualitative, point-of-care lateral flow immunoassay for the identification of AS, AC, SS/Sβ 0 thal, SC, and CC/Cβ (Nguyen-Khoa et al., 2018). The limitation HemoTypeSC has is that, it does not detect or identify the β 0 Thal phenotype. The prevalence for SCD in this study HemoTypeSC both STD POCT and DBS was 1.2% and by IEF was 0.97%. This is within range of the reported prevalence of 1.4% and 1.2% from the same environment by Nnodu et al., 2020 and the 2018 NDHS. About 10 mothers declined their babies being screened for various reasons. Averagely, we can say that the screening apathy was due to lack of proper understanding of the importance of NBS. We suggest a more educational approach to tackle this problem. Other studies with similar sample sizes, came up with similar results, for instance; A study conducted in two selected primary health care centres in Shomolu local government area (LGA) in Lagos, Nigeria involving Two hundred and ninety-one mother-infant pairs presenting for the first or second immunization visit presented similar results. In this study, the haemoglobin genotype of mother-infant pairs was determined using the HemoTypeSC rapid test kit. Confirmation for the infants’ Hb genotype was carried out using HPLC. A SCD prevalence of the infant cohort was 0.8% not up to the proposed 2% for Nigeria (Oluwole et al., 2020). In a study conducted in Democratic Republic of Congo 448 children less than 5 years of age were screened. Among this number 12.7% were homozygous (SS) (Aimé et al., 2022). In the NDHS, there were areas of SCD prevalence up to 2% but that was not representative of the whole country. 5 Conclusion HemoTypeSC test using dried blood spot is as accurate as the standard point-of-care HemoTypeSC test as can be seen from Table 5, with a Kappa value of 1.00. The use of DBS on HemoTypeSC could ensure better efficiency and cost-effectiveness in mass newborn screening for SCD. It can also provide an opportunity to leverage other public health programs such as the early infant diagnosis of HIV which utilize DBS to screen for SCD thus integrating the programs to expand SCD NBS services. Data availability statement The original contributions presented in the study are included in the article/supplementary materials, further inquiries can be directed to the corresponding authors. Ethics statement The studies involving human participants were reviewed and approved by the Federal Capital Territory Health Research Ethics Committee. Written informed consent to participate in this study was provided by the participants’ legal guardian/next of kin. Author contributions ON provided the conception, design and supervision for this work, CO carried out the study, wrote the first draft of the manuscript, CO and RC performed the statistical analysis, HI, ON, and SA read and revised the first manuscript, CO and ON revised the manuscript and responded to the reviewers comments. Funding This research was supported by the NHLBI 1UO1HL156942 Sickle Pan African Research Consortium NigEria NEtwork (SPARC-Net) and Silver Lake Research Corporation United States. Conflict of interest The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest. Publisher’s note All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher. References Aimé, A. K., Etienne, S. M., Mbongi, D., Nsonso, D., Serrao, E., Leon, T. M. M., et al. (2022). HemoTypeSC screening for sickle cell disease in the democratic republic of Congo (DRC): A case from the city of kindu. Pan Afr. Med. J. 41, 134. doi:10.11604/pamj.2022.41.134.30187 PubMed Abstract | CrossRef Full Text | Google Scholar Berger, G., Kitenge, R., Ndiaye, D. D., Ba, M. D., Adjoumani, L., Traore, H., et al. (2022). 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Family, community, and health system considerations for reducing the burden of pediatric sickle cell disease in Uganda through newborn screening. Glob. Pediatr. Health 3, 2333794X16637767. doi:10.1177/2333794X16637767 PubMed Abstract | CrossRef Full Text | Google Scholar Hsu, L., Nnodu, O. E., Brown, B. J., Tluway, F., King, S., Dogara, L. G., et al. (2018). White paper: Pathways to progress in newborn screening for sickle cell disease in sub-saharan Africa. J. Trop. Dis. Public Health 06 (02), 260. doi:10.4172/2329-891X.1000260 PubMed Abstract | CrossRef Full Text | Google Scholar Inusa, B. P., Daniel, Y., Lawson, J. O., Dada, J., Matthews, C. E., Momi S, S., et al. (2015). Sickle cell disease screening in Northern Nigeria: The co-existence of β-thalassemia inheritance. Pediat Ther. 52625 (262), 1010–41724172. Google Scholar Kanter, J., Telen, M. J., Hoppe, C., Roberts, C. L., Kim, J. S., and Yang, X. (2015). Validation of a novel point of care testing device for sickle cell disease. 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National population commission (NPC) [Nigeria] and ICF, Nigeria demographic and health Survey 2018 key indicators report, NPC and ICF, Abuja, Nigeria, and rockville. Md. U. S. A. 5. Available at: Google Scholar Nguyen-Khoa, T., Mine, L., Allaf, B., Ribeil, J. A., Remus, C., Stanislas, A., et al. (2018). Sickle SCAN™ (BioMedomics) fulfills analytical conditions for neonatal screening of sickle cell disease. Ann. Biol. Clin. 76 (4), 416–420. doi:10.1684/abc.2018.1354 PubMed Abstract | CrossRef Full Text | Google Scholar Nnodu, O. E., Sopekan, A., Nnebe-Agumadu, U., Ohiaeri, C., Adeniran, A., Shedul, G., et al. (2021). Implementing newborn screening for sickle cell disease as part of immunisation programmes in Nigeria: A feasibility study. Lancet. Haematol. 8 (10), e534e534–e540. doi:10.1016/S2352-3026(20)30143-5 CrossRef Full Text | Google Scholar Nnodu, O., Isa, H., Nwegbu, M., Ohiaeri, C., Adegoke, S., Chianumba, R., et al. (2019). HemoTypeSC, a low-cost point-of-care testing device for sickle cell disease: Promises and challenges. Blood Cells Mol. Dis. 78, 22–28. doi:10.1016/J.BCMD.2019.01.007 PubMed Abstract | CrossRef Full Text | Google Scholar Olatunya, O. S., Albuquerque, D. M., Fagbamigbe, A. F., Faboya, O. A., Ajibola, A. E., Babalola, O. A., et al. (2021). Diagnostic accuracy of HemotypeSC as a point-of-care testing device for sickle cell disease: Findings from a southwestern state in Nigeria and implications for patient care in resource-poor settings of sub-saharan Africa. Glob. Pediatr. Health 8, 2333794X211016789. doi:10.1177/2333794X211016789 PubMed Abstract | CrossRef Full Text | Google Scholar Piel, F. B., Patil, A. P., Howes, R. E., Nyangiri, O. A., Gething, P. W., Dewi, M., et al. (2013). Global epidemiology of sickle haemoglobin in neonates: A contemporary geostatistical model-based map and population estimates. 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(2019) ‘Accurate dried blood spots collection in the community using non-medically trained personnel could support scaling up routine viral load testing in resource limited settings’. doi:10.1371/journal.pone.0223573 CrossRef Full Text | Google Scholar Therrell, B. L., Lloyd-Puryear, M. A., Ohene-Frempong, K., Ware, R. E., Padilla, C. D., Ambrose, E. E., et al. (2020). Empowering newborn screening programs in African countries through establishment of an international collaborative effort. J. Community Genet. 11 (3), 253–268. doi:10.1007/s12687-020-00463-7 PubMed Abstract | CrossRef Full Text | Google Scholar Ware, R. E., de Montalembert, M., Tshilolo, L., and Abboud, M. R. (2017). Sickle cell disease. Lancet 390 (10091), 311–323. doi:10.1016/S0140-6736(17)30193-9 PubMed Abstract | CrossRef Full Text | Google Scholar Williams, T. N. (2016). Sickle cell disease in sub-saharan Africa. Hematol. Oncol. Clin. North Am. 20 (2), 343–358. doi:10.1016/j.hoc.2015.11.005 PubMed Abstract | CrossRef Full Text | Google Scholar Yang, X., Benton, M. S., Vignes, S. M., and Shevkoplyas, S. S. (2013). A simple, rapid, low-cost diagnostic test for sickle cell disease. Lab. Chip 13, 1464–1467. doi:10.1039/c3lc41302k PubMed Abstract | CrossRef Full Text | Google Scholar Keywords: sickle cell disease, dry blood spot, newborn screening, HemoTypeSC, Hb Genotype, point-of-care-test (POCT) Citation: Okeke CO, Chianumba RI, Isa H, Asala S and Nnodu OE (2022) Using dried blood spot on HemoTypeSC™, a new frontier for newborn screening for sickle cell disease in Nigeria. Front. Genet. 13:1013858. doi: 10.3389/fgene.2022.1013858 Received: 07 August 2022; Accepted: 12 October 2022; Published: 26 October 2022. Edited by: Marvin Reid, University of the West Indies, Mona, Jamaica Reviewed by: Edith Christiane Bougouma, Groupe de Recherche Action en Santé (GRAS), Burkina Faso Emily Riehm Meier, Global Blood Therapeutics, United States Copyright © 2022 Okeke, Chianumba, Isa, Asala and Nnodu. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms. Correspondence: Chinwe O. Okeke, chinwaem@gmail.com Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher. Frontiers' impact Articles published with Frontiers have received 12 million total citations Your research is the real superpower - learn how we maximise its impact through our leading community journals Explore our impact metrics Download article Download PDF ReadCube EPUB XML Share on Export citation EndNote Reference Manager Simple Text file BibTex 3,8K Total views 1,3K Downloads 15 Citations Citation numbers are available from Dimensions View article impact View altmetric score Share on Edited by Marvin Reid University of the West Indies, Mona, Jamaica Reviewed by EDITH Christiane BOUGOUMA Groupe de Recherche Action en Santé (GRAS), Burkina Faso Emily Riehm Meier Global Blood Therapeutics (United States), United States Table of contents Abstract 1 Introduction 2 Materials and methods 3 Results 4 Discussion 5 Conclusion Data availability statement Ethics statement Author contributions Funding Conflict of interest Publisher’s note References Export citation EndNote Reference Manager Simple Text file BibTex Check for updates People also looked at Nonsynonymous amino acid changes in the α-chain of complement component 5 influence longitudinal susceptibility to Plasmodium falciparum infections and severe malarial anemia in kenyan children Evans Raballah, Kristen Wilding, Samuel B. 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Perkins Establishing a database for sickle cell disease patient mapping and survival tracking: The sickle pan-african research consortium Nigeria example Obiageli Nnodu, Anazoeze Madu, Reuben Chianumba, Hezekiah Alkali Isa, Isaac Olanrewaju, Samuel Osagie, Nash Oyekanmi, Raphael Zozimus Sangeda, Annemie Stewart, Victoria Nembaware, Jack Morrice, Mario Jonas, Gaston Mazandu, Ambroise Wonkam and Olumide Owolabi Evolutionary genetics of malaria Kristan Alexander Schneider and Carola Janette Salas Blockchain Compliance by Design: Regulatory Considerations for Blockchain in Clinical Research Wendy Charles, Natalie Marler, Lauren Long and Sean Manion The types of tumor infiltrating lymphocytes are valuable for the diagnosis and prognosis of breast cancer Ying Sun and Chunyan Zhang Guidelines Author guidelines Services for authors Policies and publication ethics Editor guidelines Fee policy Explore Articles Research Topics Journals How we publish Outreach Frontiers Forum Frontiers Policy Labs Frontiers for Young Minds Frontiers Planet Prize Connect Help center Emails and alerts Contact us Submit Career opportunities Follow us © 2025 Frontiers Media S.A. 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https://drshilpagb.in/blog/life-after-hysterectomy-for-adenomyosis/
Life After Hysterectomy for Adenomyosis | Dr. Shilpa GB Dr. Shilpa G B About Us Fertility IVF Centre in Bangalore IUI Treatment in Bangalore ICSI Treatment Centre in Bangalore FET Centre in Bangalore Colposcopy Center in Bangalore Egg freezing Center in Bangalore Gynaecologic Oncology Hysteroscopy Laparoscopy PESA TESA Sperm Freezing Ultrasonography Endometriosis Surgery in Bangalore Gynaecology Uterine Fibroid Treatment Ectopic Pregnancy High-Risk Pregnancy Laparoscopic Hysterectomy Azoospermia IVF IUI ICSI FET Colposcopy Egg Freezing Gynaecologic Oncology Endometriosis Treatment Ectopic Pregnancy High-Risk Pregnancy Gynaecology Hysteroscopy Laparoscopy PESA TESA Sperm Freezing Ultrasonography Uterine Fibroid Laparoscopic Hysterectomy Azoospermia Dr. Shilpa G B Blog Contact us X Book an Appointment 7676779106 HOME>BLOG>Life after Hysterectomy for Adenomyosis Life after Hysterectomy for Adenomyosis Undergoing a hysterectomy for adenomyosis is a big decision, often taken to relieve chronic pain and discomfort. In this article, we’ll simplify what life can be like after this surgery, focusing on the physical, emotional, and lifestyle changes one might experience. We’ll also touch on medical follow-up and where you can find support and resources. Hysterectomy for Adenomyosis What is Adenomyosis? Adenomyosis is when the tissue that normally lines the uterus starts growing into the muscular wall of the uterus. This can cause heavy and painful periods and discomfort, often leading individuals to consider surgery. Types of Hysterectomy Procedures There are differenttypes of hysterectomy procedures available, each tailored to meet specific medical needs and conditions. The choice of procedure depends on factors like the severity of adenomyosis, the presence of other medical issues, and the recommendations of healthcare professionals. These procedures range from a partial hysterectomy, which involves the removal of the upper part of the uterus while leaving the cervix intact, to a total hysterectomy that includes the complete removal of both the uterus and cervix. In some cases, a radical hysterectomy may be performed, which also involves the removal of surrounding tissues, such as the ovaries and fallopian tubes, especially if cancer is a concern. Each type of hysterectomy has its own set of benefits and recovery considerations, and the decision is usually made after a thorough evaluation by a medical expert. Why Hysterectomy is Recommended A hysterectomy might be suggested when other treatments, like medication or less invasive surgeries, don’t do the trick. It’s often seen as the best way to tackle severe pain and heavy bleeding, providing patients with significant, long-term relief. This procedure involves removing the uterus, which can help alleviate symptoms and improve life quality for those dealing with chronic issues like fibroids, endometriosis, or uterine prolapse. By getting rid of the source of discomfort, a hysterectomy lets patients take back control of their health and daily activities. Physical Changes Post-Hysterectomy Immediate Recovery and Hospital Stay After the surgery, you might stay in the hospital for a few days. Recovery involves managing pain and gradually getting back on your feet. Managing Post-Surgical Pain and Discomfort Pain management is crucial. Doctors will give medication and suggest ways like light activity and rest to help ease discomfort. Hormonal Changes and Their Effects If the ovaries are removed, you may experience hormonal changes like hot flashes. Discussing hormone replacement therapy with your doctor can help manage these symptoms. Lifestyle Adjustments Resuming Daily Activities and Exercise Gradually getting back to your routine and starting gentle exercises can significantly aid recovery after surgery. It helps to enhance circulation, improve muscle strength, and elevate mood. However, it’s important to listen to your body and avoid overexertion. Always follow your doctor’s advice to prevent any complications and ensure a smooth recovery process. Dietary Changes and Nutritional Needs Eating a balanced diet rich in nutrients is crucial for supporting the body’s healing process. Incorporate a variety of foods that provide essential vitamins and minerals, such as fruits, vegetables, whole grains, and lean proteins. A diet high in protein can help repair tissues, while vitamins and minerals boost the immune system. Staying hydrated is equally important, as it helps in maintaining energy levels and facilitating recovery. Sexual Health and Intimacy After Surgery Concerns about sexual health are common after undergoing surgery. It’s essential to address these issues with patience and understanding. Open communication with your partner can alleviate anxieties and promote a supportive environment. Consulting with a Gynaecologist provider about any limitations or precautions can also be beneficial. Over time, as you continue to heal, a gradual return to intimacy can contribute to an improved emotional and physical connection. Medical Follow-up and Monitoring Regular check-ups are crucial to ensure proper recovery and to catch any complications early. During these appointments, healthcare providers can monitor your healing progress, adjust medications if necessary, and provide guidance on activities you can safely resume. It’s also important to learn to recognize signs of complications, like unusual pain or bleeding, and Consult a gynaecologist in Bangalore promptly if you notice any of these symptoms. Staying healthy post-surgery involves not just following medical advice, but also adopting a good lifestyle. This includes maintaining a balanced diet, engaging in appropriate physical activities, managing stress effectively, and keeping up with all medical appointments. These steps help to strengthen your body, support healing, and reduce the risk of further health issues Support and Resources It’s important to get information from trusted sources like healthcare providers and reputed websites. Joining online forums or local support groups can provide community support and help you feel less alone. Always reach out to healthcare professionals for personalized advice and support, especially if you’re considering hysterectomy treatment in Bangalore Conclusion Life after a hysterectomy for adenomyosis can be pain-free and fulfilling. Embrace this new chapter with patience, self-care, and positivity. Set realistic recovery goals and celebrate your achievements along the way, which can boost your confidence and optimism. By understanding and adapting to these changes, you can navigate life after a hysterectomy confidently. Book an Appointment Name Email Phone Specialties Appointment Date Appointment Time Dr. Shilpa Fertility Centre is run by Dr. Shilpa G B. Gynaecology and fertility related problems are treated at the centre. The centre provides advanced fertility solutions, such as IUI, IVF, ICSI, blastocyst, laparoscopy, sperm and egg freezing. Quick Links About Us Dr. Shilpa Blogs Contact Us Fertility IVF IUI ICSI FET Colposcopy Egg Freezing Gynaecologic Oncology Hysteroscopy Laparoscopy PESA Azoospermia TESA Sperm Freezing Ultrasonography Endometriosis Treatment Uterine Fibroid Treatment Ectopic Pregnancy High-Risk Pregnancy Laparoscopic Hysterectomy Contact Us Dr.Shilpa GB Fertility Center JP Nagar Centre,SUNDAR VILLA922, Next to Reliance Digital, 9th Cross, 22nd Main, J.P. Nagar 2nd Phase, Bengaluru – 560 078. Shanthi Shell Fertility Centre, 36th Cross Road, No. 679, 11th Main Rd, 4th Block, Jayanagar, Bengaluru, Karnataka 560011 Call Us:7676779106/+91 80560 05959 Mail Us:shilpagb@yahoo.com © 2022-24 Dr. Shilpa Fertility Centre. All Right Reserved Message usMessage us
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https://www.linkedin.com/posts/harold-kirkham-48bb6612_ieee-standard-1459-is-in-the-final-stages-activity-7310367518613729280-mria
New IEEE standard 1459 simplifies power measurements | Harold Kirkham posted on the topic | LinkedIn Agree & Join LinkedIn By clicking Continue to join or sign in, you agree to LinkedIn’s User Agreement, Privacy Policy, and Cookie Policy. Skip to main contentLinkedIn Top Content People Learning Jobs Games Join nowSign in New IEEE standard 1459 simplifies power measurements This title was summarized by AI from the post below. Harold Kirkham Semi Retired 6mo Report this post IEEE STANDARD 1459 is in the final stages of edit before being published. This is the standard that defines, for the purposes of measurement, four “power quantities:” average power, reactive power, apparent power and power factor. Of these four, only average power is a validated measurement, of a type known as “representational.” The other three are what is called “operational.” These three have provided measurement problems for over a century. Finally, there is an edition of Std 1459 that resolves the measurement problems. The Definitions section of the new standard is 28 pages long, one page shorter than the previous version, published in 2010. But its organization is greatly simplified (see picture). It no longer requires that the user know in advance how distorted or unbalanced the system is. I was chairperson of the working group that made this new revision. I am offering seminars explaining representational and operational measurements, and their impact on the new standard. Contact me if you’re interested. 14822 Comments LikeComment Share Copy LinkedIn Facebook Twitter Kamel Alboaouh Electrical Engineer - PhD 6mo Report this comment I don't see active power!! Are you referring to the average power as the active power? LikeReply 1 Reaction Richard Aspinall EE Consulting/Contract Work 6mo Report this comment I love the fact that over 150 years later we can be coming out with a cutting edge standard that tries to unambiguously define power (i have an old electricity textbook from France, 1902, that was trying the same). LikeReply1 Reaction 2 Reactions Mohammed Azharuddin Shamshuddin Hardware Architect | PhD Student | Sustainable Energy Enthusiast 6mo Report this comment Prof. Akagi has an interesting book on PQ theory for analyzing systems with distortion. He uses instantaneous power, and clarifies power factor, and true power-factor as well, it applies to balanced, unbalanced, sinusoidal, non-sinusoidal, transient and steady-state conditions of a three-phase systems, he compares it with Budenau and Fryze's approach as well! :) I'd be interested to attend the seminar to understand another perspective, thank you! :) LikeReply 1 Reaction Tripp Tucker Purveyor Of Wonderful Energy Resources 'POWER' Evangelist at 3DFS Software-Defined Electricity. Opinions are my own, RT≠endorsement 1mo Report this comment I saw your magazine article is available for public viewing just in time for Labor Day. How's your book coming along? LikeReply 1 Reaction Annie Haas Senior Technical Advisor; EPRI Gulf 6mo Report this comment Congratulations! LikeReply 1 Reaction Peter Wung Adjunct Professor in electric power related fields, including but not limited to electric machines and controls, contemporary topics in electric power. 6mo Report this comment Outstanding. LikeReply 1 Reaction Paulo F. Ribeiro EUR ING, Ph.D., IEEE Life Fellow, UMIST - The University of Manchester, UK 6mo Report this comment Very well done, Harold LikeReply 1 Reaction Umar Musa, PhD Electrical Engineer | Lecturer | Academic Researcher | 6mo Report this comment Nice work Harold Kirkham . I'll be glad to join the seminar LikeReply 1 Reaction Sanjeev Pannala Senior Researcher, National Renewable Energy Laboratory (NREL)| Secretary, IEEE Denver Section| SMIEEE| IEEE PES Subcommittees Member 6mo Report this comment Congrats Harold! Interested to join the seminar LikeReply 1 Reaction See more comments To view or add a comment, sign in 1,034 followers 37 Posts View ProfileConnect Explore content categories Career Productivity Finance Soft Skills & Emotional Intelligence Project Management Education Technology Leadership Ecommerce Show more Show less LinkedIn© 2025 About Accessibility User Agreement Privacy Policy Your California Privacy Choices Cookie Policy Copyright Policy Brand Policy Guest Controls Community Guidelines العربية (Arabic) বাংলা (Bangla) Čeština (Czech) Dansk (Danish) Deutsch (German) Ελληνικά (Greek) English (English) Español (Spanish) فارسی (Persian) Suomi (Finnish) Français (French) हिंदी (Hindi) Magyar (Hungarian) Bahasa Indonesia (Indonesian) Italiano (Italian) עברית (Hebrew) 日本語 (Japanese) 한국어 (Korean) मराठी (Marathi) Bahasa Malaysia (Malay) Nederlands (Dutch) Norsk (Norwegian) ਪੰਜਾਬੀ (Punjabi) Polski (Polish) Português (Portuguese) Română (Romanian) Русский (Russian) Svenska (Swedish) తెలుగు (Telugu) ภาษาไทย (Thai) Tagalog (Tagalog) Türkçe (Turkish) Українська (Ukrainian) Tiếng Việt (Vietnamese) 简体中文 (Chinese (Simplified)) 正體中文 (Chinese (Traditional)) Language Sign in to view more content Create your free account or sign in to continue your search Sign in Welcome back Email or phone Password Show Forgot password? 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7464
https://math.libretexts.org/Courses/Prince_Georges_Community_College/MAT_1130_Mathematical_Ideas_Mirtova_Jones_(PGCC%3A_Fall_2022)/03%3A_Probability/3.06%3A_Counting_Methods
Search x Text Color Text Size Margin Size Font Type selected template will load here Error This action is not available. 3.6: Counting Methods ( \newcommand{\vecs}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vecd}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash {#1}}} ) ( \newcommand{\id}{\mathrm{id}}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\kernel}{\mathrm{null}\,}) ( \newcommand{\range}{\mathrm{range}\,}) ( \newcommand{\RealPart}{\mathrm{Re}}) ( \newcommand{\ImaginaryPart}{\mathrm{Im}}) ( \newcommand{\Argument}{\mathrm{Arg}}) ( \newcommand{\norm}{\| #1 \|}) ( \newcommand{\inner}{\langle #1, #2 \rangle}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\id}{\mathrm{id}}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\kernel}{\mathrm{null}\,}) ( \newcommand{\range}{\mathrm{range}\,}) ( \newcommand{\RealPart}{\mathrm{Re}}) ( \newcommand{\ImaginaryPart}{\mathrm{Im}}) ( \newcommand{\Argument}{\mathrm{Arg}}) ( \newcommand{\norm}{\| #1 \|}) ( \newcommand{\inner}{\langle #1, #2 \rangle}) ( \newcommand{\Span}{\mathrm{span}}) ( \newcommand{\AA}{\unicode[.8,0]{x212B}}) ( \newcommand{\vectorA}{\vec{#1}} % arrow) ( \newcommand{\vectorAt}{\vec{\text{#1}}} % arrow) ( \newcommand{\vectorB}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vectorC}{\textbf{#1}} ) ( \newcommand{\vectorD}{\overrightarrow{#1}} ) ( \newcommand{\vectorDt}{\overrightarrow{\text{#1}}} ) ( \newcommand{\vectE}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} ) ( \newcommand{\vecs}{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } ) ( \newcommand{\vecd}{\overset{-!-!\rightharpoonup}{\vphantom{a}\smash {#1}}} ) Counting is as easy as 1-2-3, right? You already know how to count, or you wouldn’t be taking a college-level math class! Of course, but what we’ll really be investigating in this section is efficient ways of counting outcomes. Part of that task is learning what to count and what makes something different from something that we have already counted. The ultimate goal of learning to use different counting methods is so that we can integrate these methods into probability. So far the probability experiments we have worked with have had rather small numbers of outcomes. We could easily count the number of outcomes in the sample space and in the events. If there are a larger number of outcomes in the sample space, we need to develop other ways to count the outcomes without listing them all. The Fundamental Counting Principle The simplest of the counting methods is the Fundamental Counting Principle (FCP). This method is generally used when there are choices that must be made in succession and there are several options for each choice. Example 1 Gretchen is remodeling her kitchen. For her kitchen design package, she must choose a type of floor, a type of counter, and a type of sink from the options shown in the table below. Gretchen wants to know how many different kitchen designs she could make with the options she has available. Solution A tree diagram is a useful tool to visually see all the possibilities. It can also help you organize the outcomes so that you don’t miss any of them. Start the tree diagram by listing one of the options for the Floor, branching off to each of the two options for the Counter. Make sure each option of Counter repeats for each branch of the Floor. This results in (2 \times 2 = 4) Floor-Counter patterns. Next, make sure that each option for the Sink repeats for each of the 4 Floor-Counter patterns. The completed tree diagram of Gretchen's design choices is shown. We see that there are 12 possible outcomes for the kitchen design, and they are all listed on the right side of the tree diagram above. While tree diagrams provide a visual layout of choice options and outcomes, they can take time to create -- especially when there are many options for the choices or when there are lots of choices of make. A quicker way to calculate the number of final outcomes when provided different options for at each stage of choice is to multiply together the number of options at each stage. We can use multiplication to calculate the number of different design packages that Gretchen can consider: (2 \times 2 \times 3 = 12) design packages. This counting technique is called the Fundamental Counting Principle. Fundamental Counting Principle If there are (m) possible outcomes for event (A) and (n) possible outcomes for event (B), then there are a total of (m \times n) possible outcomes for event (A) followed by event (B). This principle can be generalized for three, four, or even more events. Example 2 Let’s say that a person walks into a restaurant for a three course dinner. There are four different salads, three different entrees, and two different desserts to choose from. Assuming the person wants to eat a salad, an entree, and a dessert, how many different meals are possible? Solution There are three events: choose a salad, choose an entree, and choose a dessert. According to the Fundamental Counting Principle, multiply the number of outcomes possible at each event: (\underline{4} \times \underline{3} \times \underline{2} = 24). There are 24 different meals possible. Try it Now 1 When purchasing a computer, the e-Box laptop computer offers customers several different options for screen, memory, and color as shown below. How many ways can a customer choose to personalize her selection of a new computer? 36 ways The Fundamental Counting Principle may seem like a very simple idea, but it is very powerful. Many complex counting problems can be solved using this strategy. Example 3 Some license plates in Maryland consist of three letters followed by three digits. How many license plates of this type are possible if Try it Now 2 How many 3-digit area codes can be formed where the first and last digits are odd, and digits cannot be repeated? 180 area codes Example 4 Four customers arrive at a grocery store checkout at the same time. In how many ways can the four people line up to pay for their items? Solution: We can solve this problem by thinking about making four successive choices. Any of the customers can be first so there are 4 options for the first choice. Then, there are 3 people left who can be second. Next, there are 2 customers left who can be third. Finally, there is only 1 person left to be last in the line. Using the Fundamental Counting Principle there are (\underline{4} \times \underline{3} \times \underline{2} \times \underline{1} = 24) ways for the four customers to line up. The multiplication pattern above appears so often in counting that it has its own name, called a factorial, and its own symbol, which is '(!)'. We say “four factorial” and we write "(4!)". Factorial If (n) is a counting number, then (n) factorial ((n!)) is defined as (n! = n(n-1)(n-2)(n-3)...(2)(1)) (0! = 1) Example 5 Evaluate (5!), (8!), and (12!). (\begin{aligned} 5! &= 5 \times 4 \times 3 \times 2 \times 1 \[4pt] &= 120 \end{aligned}) (\begin{aligned}8! &= 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \[4pt] &= 40,320 \end{aligned}) (\begin{aligned} 12! &= 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \[4pt] &= 479,001,600 \end{aligned}) Factorials get very large, very fast. It is not pleasant to type so many factors when (n) is large. Most scientific calculators have a factorial command. You can find this command on the TI calculator as follows: A calculator will express large values of factorials in scientific notation. As shown below, the TI calculator finds (40!) to be almost (8.16 \times 10^{47}). Examples 4 and 5 illustrate a type of common counting problem and method called a permutation. A permutation is an ordered arrangement of objects. The key to recognizing a permutation is that it is "ordered." For example, using only the three letters C, A, and T, there are six different three-letter permutations or sequences that we can make: ACT, ATC, CAT, CTA, TAC, and TCA. If we add a fourth letter to the list, say S, then there are exactly 24 different four-letter permutations: ACST CAST SACT TACS ACTS CATS SATC TASC ASCT CSAT SCAT TCAS ASTC CSTA SCTA TCSA ATCS CTAS STAC TSAC ATSC CTSA STCA TSCA Counting the number of ordered arrangements of the four letters C, A, T, S is the same problem as counting the number of ways four grocery customers can line up at the check out stand. Therefore, when counting how many ways a group of objects can be ordered, we use the factorial of the number of objects to be ordered. Permutation A permutation is an arrangement of a set of objects without repetition where a different order of the same set of objects counts as a different arrangement. The number of permutations of (n) different objects, taken altogether, is (n!). Example 6 The school orchestra is planning to play eight pieces of music at their next concert. In how many different ways can the pieces of music be sequenced in the program? Solution This is a permutation because the orchestra is arranging all 8 songs in an order to make the program. We can use the Fundamental Counting Principle or a factorial to count the number of ordered arrangements: (\underline{8} \cdot \underline{7} \cdot \underline{6} \cdot \underline{5} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2} \cdot \underline{1}=40,320) or (8!=40,320) There are 40,320 different ways of sequencing the songs in the program. Try it Now 3 There are 6 DVD's to be placed on a shelf. In how many ways can they be placed on the shelf from left to right? (6!=720) ways But what if the orchestra in the previous example has time to only play 5 of the 8 musical pieces. How many ways could the orchestra design the program of music? This is still a permutation but a slightly different question. This type of problem will be explored next. Permutations We now consider permutations of a set of objects taken from a larger set. Example 7 The school orchestra has learned 8 pieces of music to play at their next concert. However, due to time restraints, they can only chose 5 pieces to play. In how many different ways can the pieces of music be chosen and sequenced in the program? Solution We can think of this as making 5 choices in a row: There are 8 options for the first song choice, 7 options for the second song choice, 6 options for the third song choice, 5 options for the fourth song choice, and 4 options for the fifth song choice. We can use the Fundamental Counting Principle to calculate the number of ways to choose and arrange the five pieces of music: (\underline{8} \times \underline{7} \times \underline{6} \times \underline{5} \times \underline{4} = 6,720) ways. We say that the number of permutations of 8 songs taken 5 at a time is 6,720. Notice that in this example of counting the number of ways to select and order only 5 of the 8 songs, we could not quite use (8!) because not all 8 musical pieces are being used. Notice, however, that (8 \times 7 \times 6 \times 5 \times 4 = \dfrac{8!}{3!} = \dfrac{8!}{(8-5)!}). We can generalize the preceding observation to create a formula for counting the number of permutations of (r) objects from a group of (n) objects. Permutation Formula The number of permutations of (n) objects taken (r) at a time is given by the formula​​​ (_nP_r =\dfrac{n !}{(n-r) !}). It should be emphasized that a permutation problem is nothing more than a special case of a Fundamental Counting Principle problem. Using either strategy to count the number of permutations will arrive at the same answer. Example 8 There are 10 cars in a race. In how many ways can three cars be awarded 1st, 2nd, and 3rd place? Solution The order in which the cars finish is important to consider when counting the number of ways the cars can finish. We say this is a "permutation of 10 choose 3" and can write it symbolically as (_{10}P_{3}). Using the Fundamental Counting Principle, there are (\underline{10} \cdot \underline{9} \cdot \underline{8} = 720) ways for cars to finish in the top three places. Alternately, using the permutation formula, (_{10}P_3=\dfrac{10 !}{(10-3) !}=\dfrac{10 !}{7 !}=\dfrac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} =\dfrac{10 \cdot 9 \cdot 8 \cdot \cancel{7} \cdot \cancel{6} \cdot \cancel{5} \cdot \cancel{4} \cdot \cancel{3} \cdot \cancel{2} \cdot \cancel{1}}{\cancel{7} \cdot \cancel{6} \cdot \cancel{5} \cdot \cancel{4} \cdot \cancel{3} \cdot \cancel{2} \cdot \cancel{1}} = 10 \cdot 9 \cdot 8 = 720 ) It should be noted that most calculators have a permutation command. On a TI calculator, it is found in the same menu as the factorial. To evaluate (_{10}P_3) on the TI calculator, type the value of (n), go to the MATH menu and move right to the PRB sub-menu, select the (_nP_r) command, type the value of (r), and press ENTER. Here is the sequence of screens. Example 9 There are 4 hooks in a row on a wall to hang some pictures. You have 7 pictures to display. Find the number of ways can you choose and arrange 4 pictures on the hooks from the group of 7 pictures. Solution The order in which the pictures are arranged is important. Choosing the same group of 4 pictures but placing them in a different order creates a different arrangement. This is a permutation of 7 choose 4, or (_{7}P_4). Using the Fundamental Counting Principle, there are 4 choices to make with 7 options for the first choice, 6 options for the second choice, 5 options for the third choice, and 4 options for the fourth choice: (\underline{7} \cdot \underline{6} \cdot \underline{5} \cdot \underline{4} =840) ways Using the permutation formula, (_{7}P_4=\dfrac{n !}{(n-r) !}=\dfrac{7 !}{(7-4) !}=\dfrac{7 !}{3 !}=\dfrac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1}=\dfrac{7 \cdot 6 \cdot 5 \cdot 4 \cdot \cancel{3} \cdot \cancel{2} \cdot \cancel{1}}{\cancel{3} \cdot \cancel{2} \cdot \cancel{1}}=7 \cdot 6 \cdot 5 \cdot 4=840) There are 840 different ways the to choose and arrange 4 pictures on the hooks. Example 10 The Volunteer Club has 18 members. An election is held to choose president, vice-president, and secretary. In how many ways can the three officers be chosen? Solution The order in which the officers are chosen matters. Choosing the same three people but assigning them to a different role would make a different way of choosing the officers. That is, A as president, B as vice-president, and C as secretary is different from B as president, C and vice-president, and A as secretary. This is a permutation of 18 choose 3, or (_{18}P_3). Using the Fundamental Counting Principle, there are 3 choices to make with 18 options for the first choice, 17 options for the second choice, and 16 options for the third choice: (\begin{array} {cccccc} {\underline{18}}&{\cdot}&{\underline{17}}&{\cdot}&{\underline{16}}&{ = 4896}\ {\text{Pres.}}&{}&{\text{V.P.}}&{}&{\text{Sec.}}&{} \end{array}) Using the permutation formula, (_{18}P_3=\dfrac{n !}{(n-r) !}=\dfrac{18 !}{(18-3) !}=\dfrac{18 !}{15 !}=\dfrac{18 \cdot 17 \cdot 16 \cdot \cancel{15!}}{\cancel{15!}}=18 \cdot 17 \cdot 16=4,896) To make the simplification a bit shorter, several of the factors of (18!) in the numerator cancel with (15!) in the denominator. There are 4,896 different ways the three officers can be chosen. Example 12 How many 5 character passwords can be made using the letters A through Z if letters can be used only once? Solution As we all know, the order in which you type the letters in your password matters! Using the correct letters in the password but in the wrong order won't unlock your account. Since a different order of the same five letters makes a different password, this is a permutation of 18 choose 3, or (_{18}P_3). Using the Fundamental Counting Principle, (\underline{26} \cdot \underline{25} \cdot \underline{24} \cdot \underline{23} \cdot \underline{22} =7,893,600) Using the permutation formula, (_{26}P_5=\dfrac{n !}{(n-r) !}=\dfrac{26 !}{(26-5) !}=\dfrac{26 !}{21 !}=\dfrac{26 \cdot 25 \cdot 24 \cdot 23 \cdot 22 \cdot \cancel{21!}}{\cancel{21 !}}=26 \cdot 25 \cdot 24 \cdot 23 \cdot 22=7,893,600) There are 7,893,600 different passwords made of only 5 letters of the alphabet. Try it Now 4 Twelve actresses are available to be cast in a play that has five female roles. In how many ways could the 5 roles be cast? 95,040 ways Combinations We have considered the situation where we chose (r) items from a group of (n) items where the order of selection is important in distinguishing one group from another. We now consider a similar situation in which the order of selection is not important. A collection of items, in no particular order, is called a combination. Using our language of sets, a combination is a subset of a set of objects. For example, suppose that in a group of five students — Andy (A), Barry (B), Cheryl (C), Darren (D), and Ellen (E) — three students are to be selected to make a team. Each of the possible three-member teams is a combination. How many such combinations are there? We can answer this question by using our knowledge of permutations and the Fundamental Counting Principle. If order did matter in the selection of the three students for the team, permutations would be counted as (_5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!}=60) teams. But, since order doesn’t matter in this case, this number of permutations counts more teams than there should be. We need to divide out the number of teams that have been repeated but are in a different order. When order of team members makes no difference, the permutations ABC, ACB, BAC, BCA, CAB, and CBA are all really the same team, ABC. That is, there are six three-person permutations for each team of three people. This is consistent with the fact that three objects can be rearranged in (3!=6) different ways. To compute the number of possible combinations of three students chosen from a group of five students, we can divide the number of permutations by the number of repeated teams. So, one technique for counting the number of combinations of 5 objects chosen 3 at a time is (_5C_3 = \dfrac{_5P_3}{3!} = \dfrac{60}{6} = 10 ) teams, which could also be written as (_5C_3 = \dfrac{_5P_3}{3!} = \dfrac{5!}{(5-3)!\;3!} = \dfrac{5!}{2!\;3!} = = \dfrac{5 \times 4 \times \cancel{3} \times \cancel{2} \times \cancel{1}}{(2 \times 1) \times (\cancel{3} \times \cancel{2} \times \cancel{1})}=10 ) teams. We can generalize the preceding observations to write a formula for the number of combinations of (r) objects selected from a group of (n) objects. Combination Formula A combination is a selection of a set of objects without repetition in which the order of selection does not matter. The number of combinations of (n) objects is taken (r) at a time is given by the formula​​​ (_nC_r =\dfrac{_nP_r}{r!} = \dfrac{n !}{(n-r)!\;r!}). Example 13 A college class has a reading list of eight books. A student must choose and read five of the books before the end of the course. In how many ways can the student choose five books to read? Solution The order of selecting the books is not important, only which books are read. That is, as long as the same 5 books are selected it is the same choice no matter which order they are selected. We say this is a "combination of 8 choose 5" and can write it symbolically as (_{8}C_{5}). Using the formula for counting combinations and simplifying, (C(8,5)=\dfrac{n !}{(n-r)! \;r!}=\dfrac{8 !}{(8-5)! \;5!}=\dfrac{8 !}{3 ! \times 5 !}= \dfrac{8 \times 7 \times 6 \times 5! }{3! \times 5!} = \dfrac{8 \times 7 \times 6 \times \cancel{5!} }{3! \times \cancel{5!}} = \dfrac{8 \times 7 \times 6 }{ 3 \times 2 \times 1} = \dfrac{8 \times 7 \times 6 }{6}= \dfrac{8 \times 7 \times \cancel{6} }{ \cancel{6}} = 8 \times 7 = 56) There are 56 ways to choose five of the books to read. As with permutations, most calculators have a combination command. On the TI calculator, it is found in the same menu as factorial and permutation. To find (_{8}C_5) on the TI calculator, type the value of (n), go to the MATH menu and move right to the PRB sub-menu, select the (_nC_r) command, type the value of (r), and press ENTER. Here is the sequence of screens. Example 14 A child wants to pick three pieces of candy to take in her school lunch box. If there are 13 pieces of candy to choose from, how many ways can she pick the three pieces? Solution This is a combination because it does not matter in what order the candy is chosen. The same 3 pieces selected in any order gives the child the same candies. This is a combination of 13 choose 3, or (_{13}C_3). Using the combination formula, ( _{13}C_3 =\dfrac{n !}{(n-r)! \; r!} =\dfrac{13 !}{(13-3)! \; 3!} = \dfrac{13 !}{10! \times 3!} = \dfrac{13 \times 12 \times 11 \times 10! }{10! \times 3!} = \dfrac{13 \times 12 \times 11 \times \cancel{10!} }{\cancel{10!} \times 3!} = \dfrac{13 \times 12 \times 11 }{ 3 \times 2 \times 1} = \dfrac{1716 }{6}= 286 ) There are 286 ways to choose the three pieces of candy to pack in her lunch. Example 15 The Volunteer Club has 18 members. A committee of members will be selected to plan the annual food drive. How many different 3-person committees can be selected? Solution Unlike selecting members for officers where each officer serves a different role, the order in which the members are chosen for a committee is not meaningful. Choosing the same group of three people but in a different order results in the same committee of three people. This is a combination of 18 choose 3, or (_{18}C_3). (_{18}C_3 =\dfrac{n !}{(n-r)! \; r!} =\dfrac{18 !}{(18-3)! \; 3!} =\dfrac{18 !}{15 ! \times 3 !} = \dfrac{18 \times 17 \times 16 \times 15! }{15! \times 3!} = \dfrac{18 \times 17 \times 16 \times \cancel{15!}} {\cancel{15!} \times 3!} = \dfrac{18 \times 17 \times 16 }{ 3 \times 2 \times 1} = \dfrac{4896 }{6}= 816 ) You could have also used the result from Example 10: (_{18}C_3 = \dfrac{_{18}P_3}{3!}=\dfrac{4896}{6}=816). There are 816 ways to choose a committee of three members. Try it Now 5 You have 4 extra tickets to the Nationals game and 9 of your friends want to go. How many ways can you select a group of friends to join you at the game? 126 ways Simplifying permutations and combinations by hand can be tedious for large quantities so most of the time we will want to use technology. The difficulty for most people is knowing whether a problem calls for a permutation, a combination, or only the Fundamental Counting Principle. The table gives a quick summary: Distinguishing between Fundamental Counting Principle, Permutation, and Combination | | | --- | | Fundamental Counting Principle | Counts the number of ways for event (A) followed by event (B) when event (A) has (m) outcomes and event (B) has (n) outcomes: (m \times n) Can be used with a sequence of more than 2 events. May be used with or without repetitions. Particularly useful when there are specific placement requirements or conditions to meet (e.g. first digit must be odd.) | | Permutation | Counts the number of ways to select (r) items from a group of (n) items when the order of items is important (_nP_r = \dfrac{n!}{(n-r)!}) ABC is different than BAC. Items cannot be repeated. May also be solved using the Fundamental Counting Principle. | | Combination | Counts the number of ways to select (r) items from a group of (n) items when the order of items is not important (_nC_r = \dfrac{n!}{(n-r)!\; r!}) ABC is the same as BAC. Items cannot be repeated. May also be solved by dividing the number of permutations by (r!) | Fundamental Counting Principle Counts the number of ways for event (A) followed by event (B) when event (A) has (m) outcomes and event (B) has (n) outcomes: (m \times n) Permutation Counts the number of ways to select (r) items from a group of (n) items when the order of items is important (_nP_r = \dfrac{n!}{(n-r)!}) Combination Counts the number of ways to select (r) items from a group of (n) items when the order of items is not important (_nC_r = \dfrac{n!}{(n-r)!\; r!}) In the examples that follow, we concentrate on identifying which method to use and calculate with technology. Example 16 A jury pool consists of 20 people. How many different 12-person juries can be selected from the jury pool? Solution Choosing the same 12 people in a different order for the jury would not result in a different outcome. Order does not make a difference so this is a combination: (_{20}C_{12} = 125,970) juries. Example 17 A swimming event has 16 contestants. The swimmers with the five fastest speeds will be listed, in the order of their speed, on the leader board. How many ways are there for the names of five swimmers to be listed? Solution Listing the same 5 names in a different order on the leader board would be a different result to the race. Order is important so this is a permutation: (_{16}P_5 = 524,160) ways. Example 18 You are completing a four-question survey on your experience at Taco Bell. You can answer Below Average, Average, or Above Average for each question. In how many different ways could you answer this survey? Solution Recall that when items can be selected more than once you cannot use the permutation or combination formulas. It is possible to respond "Average" to more than one of the four questions. There are four choices to be made, and there are three options for each choice. Use the Fundamental Counting Principle: (\underline{3} \times \underline{3} \times \underline{3} \times \underline{3} = 81) ways. Example 19 When playing poker, players are dealt 5 cards from a regular deck of cards. How many different hands of poker could a player be dealt? Solution Receiving the same five cards but in a different order would not mean that you had a different set of cards. Order does not make a difference so this is a combination: (_{52}C_{5} = 2,598,960) hands. Try it Now 6 Decide whether the scenario requires permutations, combinations, or the Fundamental Counting Principle. Then, use technology to compute the answer. Using More than One Method Sometimes a counting problem may require more than one counting method. For example, you may need to compute combinations and use the Fundamental Counting Principle together. Let's look at a relatively simple but common example. Example 20 A sandwich shop offers a special: choose exactly one kind of bread, one kind of protein, and three toppings from the menu below and get a special price. How many ways can a customer select menu items for the special? Solution This problem looks very similar to examples where we used the Fundamental Counting Principle. However, in those problems, we chose only one option for each stage of choice. Here, we are choosing one type of bread, one type of protein, but three types of vegetable toppings. The Fundamental Counting Principle can be applied by multiplying the number of ways a bread can be selected (2), the number of ways a protein can be selected (4), and the number of ways 3 vegetables can be selected from 5 vegetables ((_5C_3=10)): (\underline{2} \times \underline{4} \times \underline{_5C_3} = 2 \times 4 \times 10 = 80). So, there are 80 different ways of selecting menu items for he special. Example 21 Brenda will choose 5 movies to rent over the weekend, and she has decided to rent 3 science fiction movies and 2 comedies. She can choose from 6 science fiction movies and 4 comedies. How many different ways can Brenda choose the group of 5 movies? Solution There are two different stages to selecting the group of movies. First, compute how many ways Brenda can select 3 science fiction movies from a group of 6 science fiction movies using (_6C_3=\frac{6!}{(6-3)! \; 3!} = 20). Then, compute how many ways Brenda can choose 2 comedies from the group of 4 comedies using (_4C_2= \frac{4!}{(4-2)! \; 2!} = 6). Each group of science fiction movies can be paired with each group of the comedies so we use the Fundamental Counting Principle to find how many different groups of 3 science fiction movies and 2 comedies can be selected together: (_6C_3 \times _4C_2 = 20 \times 6 = 120). There are 120 ways for Brenda to choose her group of 5 movies. Try it Now 7 An art gallery has a total of 11 paintings by a certain artist. Of these paintings, 5 are oil paintings and 6 are watercolor paintings. The art gallery will display a special exhibition of this artist’s work but is restricted to showing only 7 paintings. Calculate the number of ways in which the 7 paintings can be selected for the exhibition if it includes 3 oil paintings and 4 watercolor paintings. 150 ways In the next section we will apply these counting methods when we return to calculating theoretical probability. 3.6: Counting Methods is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. Recommended articles The LibreTexts libraries are Powered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. 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Art of Problem Solving Constructive counting - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Constructive counting Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Constructive counting In combinatorics, constructive counting is a counting technique that involves constructing an item belonging to a set. Along with the construction, one counts the total possibilities of each step and assembles these to enumerate the full set. Along with casework and complementary counting, constructive counting is among the most fundamental techniques in counting. Familiarity with constructive counting is essential in combinatorics, especially in intermediate competitions. Contents [hide] 1 Introductory Examples 1.1 Example 1 1.2 Example 2 1.3 Example 3 1.4 Example 4 1.5 Example 5 2 Intermediate Examples 2.1 Example 1 3 More Problems 4 Resources 5 See also Introductory Examples Example 1 How many four-digit numbers are there? Solution: We can construct a four-digit by picking the first digit, then the second, and so on until the fourth. The first digit can be any number from one to nine — zero excluded, or else it ceases to have four digits — so it has possible choices. The other three digits can be any number from zero to nine (total 10 digits), so they all have possibilities. We multiply the possibilities for each digit to arrive at our answer: . Constructive counting is a simple concept, more so than its definition might lead one to think. All we did was think about constructing a four-digit number by choosing its digits, compute the possible numbers each digit can be, and multiply the resulting numbers. This is a problem where constructive counting is not the simplest way to proceed. This next example is one where constructive counting is essential: Example 2 How many lists of seven numbers are there such that each entry is between and inclusive and no two consecutive entries are equal? Solution: We can model this situation as a row of 7 boxes, like this: which we must populate with numbers between and including and ; the key restriction here is that no two boxes right next to each other can have the same number. The first digit can be any number from to , of which there are such options. The second can be any of to too, with the sole exception of the previous digit. Regardless of whatever the first digit is, we know that it removes one option, so there are options for the second digit. The exact same logic applies to the third digit; it can be any of the digits except the one before it, so it has options. This continues until the seventh digit, which means that digits to all have options. Hence, our answer is , as desired. These two examples use options specifically based on digits, but this isn't the entire picture of constructive counting. The following example uses constructive counting in a different context. Example 3 How many permutations of are there? Solution: We can model the question as a row of seven boxes, like this: which we have to populate with s and s. Using constructive counting is an idea, but there are multiple ways one might proceed with the construction. If we were to go like before and break the problem down by each box, we'd get a fairly messy solution. Instead, one might think to break it down by first placing the s separately, then placing the s. Starting with the s, we must choose the boxes of their placement; because all the s are indistinguishable, this is given by , where is a combination. One example among many placements of the s is For the s, their position is actually predetermined by choosing the s; the only place the three s can go is in the three empty boxes, so we don't have to account for them after choosing the s. Thus, there are different permutations of , as required. . Like in this problem, there are sometimes multiple independent ways to construct a set. In others, however, an alternative method is not apparent, as with the next example: Example 4 2001 AMC 12 Problem 16: A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe? Solution: Note that each leg has one designated shoe and a designated sock; each leg's shoe and sock belong to only that leg. The question is then only asking about the order in time in which it puts on all 16 socks and shoes. We can model the different orders as 16 boxes, where each box is populated with a certain sock or shoe, like this: Breaking up the problem by each box leads to a dead end. Instead, we can use a similar approach to example three to solve this problem — we can first choose two boxes in which each leg's sock-shoe pair is located, then we permute them. For the first leg, the location of its sock and shoe's two boxes are given by ; but on permuting them, we know that the sock appears first in the list, which implies that only one permutation exists. So, there are just different places in which the first leg's sock and shoe can be located in the 16 boxes. By similar logic, the second leg has places in the boxes, the third has , and so on. The final answer is then the product of the leg's choices, which is Thus, our final answer is . . This problem was much more challenging than the others mentioned thus far, but it's a lovely illustration of just how effective a cleverly chosen construction can be to counting problems. Example 5 2004 AIME I Problem 6: An integer is called snakelike if its decimal representation satisfies if is odd and if is even. How many snakelike integers between 1000 and 9999 have four distinct digits? Solution: We construct the set of snakelike integers. All the recursive requirement is saying is that the second digit must be greater than the first and third, and the fourth must be greater than the third. But before we start construction, we must divide our investigation into several cases, based on whether we allow zeroes Case 1: Snakelike integers with no zero. First, we choose the four integers. They must be between and inclusive, and no digit can be repeated, so the combination that describes this is . Next, we find out how many permutations of these digits keep the number snakelike. For simplicity, consider the case of , , , and . The total possibilities are This applies to all snakelike integers with no zero, which means there arrangements that keep the number snakelike. Thus we have snakelike integers with four distinct digits and no zero. Case 2: Snakelike integers with one zero. First, we choose the other three integers, which by similar logic above is . Next, we permute these digits. Without loss of generality, let our other three digits be , , and . The total possibilities are Hence, there are snakelike integers with four distinct digits and one zero. It's easy to see that introducing a second zero would mandate them being the first and third digits. However, this breaks our requirement that our integers must be between and , so there are no four-digit snakelike integers with two or more zeroes. Thus, our total is snakelike integers with four distinct digits, as desired. . It's worth noting that as problem-solving ability becomes more advanced, there are fewer problems that can be solved by constructive counting alone; the list of examples terminates at a late-introductory level consequentially. At the same time, there are many, many more problems at a higher level where constructive counting is a crucial intermediate step, combined with other counting strategies to reach an answer. Intermediate Examples These problems are more advanced than the introductory examples, requiring greater creativity to solve. Example 1 Russia 1998: A 10-digit number is said to be interesting if its digits are all distinct and it is a multiple of 11111. How many interesting numbers are there? Solution: There doesn't seem to be a useful pattern in multiples of 11111 And even if there was one, it's not clear how to mediate it with the distinct digits condition. Thus, we look for additional restrictions that will help us count. Let be any interesting number. Because has ten distinct digits, its digits must be an ordered arrangement of to used exactly once. From here, note that the sum of its digits, , is divisible by . This implies through divisibility rules that is a multiple of ; therefore, is a multiple of . An idea from here is to utilize the special properties of — namely, that . The compells us to express as for some five-digit numbers and ; doing so gives that Both sides of this equation must be divisible by , which implies that is divisible by . It's easy to verify that even without this condition, the maximum of is , which is less than . Therefore, must be equal to . This also interacts with the distinct digits property; namely, that the corresponding digits of and add to . For example , where . Thus, corresponding digits must come in pairs — which now enables us to construct and . We first place these pairs in (which places it in ), of which they can be in any order. There are then options for the pairs' positions. The only thing left is which number in each pair goes to and which to ; there are ways we can divvy a pair up, which means there are total possibilities. But wait! This construction counts numbers that start with zero — something that violates the 10-digit condition — as being interesting numbers. Note that in our count, each digit is equally likely to be first; thus, of our numbers start with , and only of our count keeps the number ten-digit. Putting this all together, there are interesting numbers. More Problems 2005 AIME I Problems/Problem 5 Resources AoPS Constructive Counting Part 1 AoPS Casework Counting Part 1 See also Casework Complementary counting Overcounting Retrieved from " Categories: Combinatorics Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
7466
https://www.intmath.com/blog/mathematics/determining-velocity-with-time-and-change-in-acceleration-12486
Skip to main content Determining Velocity with Time and Change in Acceleration By Kathleen Cantor, 30 Sep 2020 Every object experiencing an acceleration must have a velocity. This is explained by a branch of physics which is called dynamics. It's an aspect of physics where you study the motion of an object and the forces acting on them. We can't talk about velocity without talking about speed. By definition, speed is the rate of change of distance with time, while the instrument used to measure the velocity of a moving object is called a speedometer. It measures in kilometer per hour, miles per hour, or feet per hour. Determining Velocity With Time And Change In Acceleration In physics, speed and velocity are often used intermittently. And it is easy to convert from speed to velocity. The unit of speed is kilometer per hour (km/hr.), while that of velocity is meters per second (m/s). To convert from speed to velocity, you can use the formula below (Km/hr. x 1000)/ (60 x 60) = m/s For example, to convert 36 km/hr. to m/s (36 x 1000)/ (60 x 60) = 10 m/s Types of Speed Constant/ Uniform Speed:If the rate of change of distance with time is constant throughout a journey, the speed is said to be uniform or constant. Average Speed:The average speed is the ratio of the total distance traveled to the total time taken throughout a journey. Essentially, the average speed is (total distance travelled)/ (total time taken). For example, if a car covers 150 km in 2 hrs, the average speed is (150)/ (2) = 75 km/hr. Types of Velocity Velocity can be defined as the rate of change of displacement with time. Therefore, velocity = (displacement)/ (time) = m/s Initial Velocity:This is the velocity of an object before there was an increased acceleration or change in velocity, and it is denoted with U. Final Velocity:This is the velocity of an object after there was an increased acceleration or change in velocity, and it is denoted with V. Uniform Velocity If the rate of change of displacement with time is constant throughout a journey, the body is said to be moving with a uniform velocity. The change in velocity the velocity of an object is simply the final velocity minus the initial velocity. This change in velocity is also known as acceleration. For example, Change in Velocity = Final Velocity – Initial Velocity Acceleration Acceleration is the rate of change of velocity with time. When an object increases its velocity with time, it's said to accelerate. Therefore, acceleration (a) = (Velocity)/ (Time) = (Change in Velocity)/ (Change in Time) = m/s2 Uniform Acceleration If the rate of change of velocity with time is constant, the acceleration is said to be uniform. If an object’s velocity is decreased, then the acceleration of the object will also decrease with time. When the velocity of an object decreases with time, the process is called deceleration, retardation, or negative acceleration. Example 1 Calculate the acceleration of a train which travels at 36 km/hr and accelerates uniformly to 108 km/hr in 10 sec. Solution The initial velocity = (36 x 1000)/ (60 x 60) = 10 m/s Final Velocity = (108 x 100)/ (60 x 60) = 30 m/s Acceleration (a) = (change in velocity)/ (change in Time) = (30 – 10)/ (10 – 0) = 2 m/s Example 2 A car starting from rest is uniformly accelerated so that its velocity in 5 sec. is 36 km/hr. A break is then applied for it to stop in 4 sec. Find (a) the acceleration (b) the retardation. Solution Since the car is starting from rest, the initial velocity (u) = 0 The final velocity (v) = 36 km/hr. = (36 x 1000)/ (60 x 60) = 10 m/s, time (t) = 5 sec. The acceleration (a) = (Final velocity – Initial velocity)/ (change in Time) Therefore, (a) = (10 – 0)/ (5 – 0) = 2 m/s The retardation = negative acceleration, v = 0 (since the car has come to a stop), u = 10 m/s, t = 4sec. Therefore, (a) = (0 – 10)/ (4 – 0) = -10/4 = -2.5 m/s2 Change In Acceleration According to Newton’s second law which states that when a force acts on an object, it causes the object to accelerate (change the object’s velocity) at a constant rate. It means that when a force is applied to an object at rest, it will cause it to move in the direction of the force. But if the object is already in motion, it will speed up, slow down or change the direction of movement of the object. From Newton’s second law F = Ma, where F = force acting on the object, M= mass of the object, and a= acceleration caused by force. If you make ‘a’ to be the subject of the formula, then a = F/M Therefore, if we have a situation in which an object of mass (M) is under the influence of a force ‘F’ with an acceleration (a). If the force is increased, then there will be a positive change in acceleration (a), but if the force is decreased, there will be a negative change in acceleration. As a result, change in acceleration = (F2 – F1)/M Sample 3 If an object of mass 5 kg is acted upon by a force of 50N to make it move at a velocity of 5 m/s. Find the change in acceleration of the object if the force is increased to 150N. Solution F = Ma, F1 = 50N, F2 = 150N, M = 5kg Change in acceleration (a) = (150 – 50)/ (5) = 100/5 = 20 m/s2 Wrapping Up Calculating the change in velocity with time and change in acceleration of an object is not as hard as you think. You just need to apply the knowledge you have gained from this topic. Make sure that all the variables are in their standard units. Be the first to comment below. Related posts: Calculating Acceleration with Force and Mass As is usually the case in mathematics and physics, formulas and experiments usually begin with... Finding Dimensional Formula For Acceleration Physical quantities are used to quantify the properties of a system. To give meaning to... Explore the slope of the sin curve Use an interactive graph to explore how the slope of sine x changes as x... IntMath Newsletter: Angular velocity, face numbers, WIRIS In this Newsletter: 0. Ad blockers 1. Linear & angular velocity applet 2. Your face... Finding Derivative, Second Derivative, and Curvature The mathematical study of calculus requires a deep understanding of a fundamental concept: derivatives. For... Posted in Mathematics category - 30 Sep 2020 [Permalink] Leave a comment Tips Ten Ways to Survive the Math Blues How to understand math formulas How to learn math formulas How to make math class interesting? 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7467
https://en.wikipedia.org/wiki/Unary_numeral_system
Jump to content Unary numeral system العربية 閩南語 / Bn-lm-gí Беларуская Беларуская (тарашкевіца) Català Čeština Dansk Deutsch Español Esperanto فارسی Français 한국어 Ido Italiano עברית Kreyòl ayisyen Magyar Nederlands 日本語 Polski Português Romnă Русский Slovenčina Slovenščina Svenska தமிழ் ไทย Türkçe Українська Tiếng Việt 文言 粵語 中文 Edit links From Wikipedia, the free encyclopedia Base-1 numeral system | | | Part of a series on | | Numeral systems | | Place-value notation | | | Hindu–Arabic numerals Western Arabic Eastern Arabic --- Bengali Devanagari Gujarati Gurmukhi Odia Sinhala Tamil Malayalam Telugu Kannada Dzongkha --- Tibetan Balinese Burmese Javanese Khmer Lao Mongolian Sundanese Thai | | East Asian systems Contemporary Chinese + Hokkien + Suzhou Japanese Korean Vietnamese --- Historic Counting rods Tangut | | Other systems History --- Ancient Babylonian --- Post-classical Cistercian Mayan Muisca Pentadic Quipu Rumi --- Contemporary Cherokee Kaktovik (Iñupiaq) | | By radix/base Common radices/bases 2 3 4 5 6 8 10 12 16 20 60 --- Non-standard radices/bases Bijective (1) Signed-digit (balanced ternary) Mixed (factorial) Negative Complex (2i) Non-integer (φ) Asymmetric | | | Sign-value notation Non-alphabetic Aegean Attic Aztec Brahmi Chuvash Egyptian Etruscan Kharosthi Prehistoric counting Proto-cuneiform Roman Tally marks --- Alphabetic Abjad Armenian Alphasyllabic + Akṣarapallī + Āryabhaṭa + Kaṭapayādi Coptic Cyrillic Geʽez Georgian Glagolitic Greek Hebrew | | List of numeral systems | | v t e | The unary numeral system is the simplest numeral system to represent natural numbers: to represent a number N, a symbol representing 1 is repeated N times. In the unary system, the number 0 (zero) is represented by the empty string, that is, the absence of a symbol. Numbers 1, 2, 3, 4, 5, 6, ... are represented in unary as 1, 11, 111, 1111, 11111, 111111, ... Unary is a bijective numeral system. However, although it has sometimes been described as "base 1", it differs in some important ways from positional notations, in which the value of a digit depends on its position within a number. For instance, the unary form of a number can be exponentially longer than its representation in other bases. The use of tally marks in counting is an application of the unary numeral system. For example, using the tally mark | (𝍷), the number 3 is represented as |||. In East Asian cultures, the number 3 is represented as 三, a character drawn with three strokes. (One and two are represented similarly.) In China and Japan, the character 正, drawn with 5 strokes, is sometimes used to represent 5 as a tally. Unary numbers should be distinguished from repunits, which are also written as sequences of ones but have their usual decimal numerical interpretation. Operations [edit] Addition and subtraction are particularly simple in the unary system, as they involve little more than string concatenation. The Hamming weight or population count operation that counts the number of nonzero bits in a sequence of binary values may also be interpreted as a conversion from unary to binary numbers. However, multiplication is more cumbersome and has often been used as a test case for the design of Turing machines. Complexity [edit] Compared to standard positional numeral systems, the unary system is inconvenient and hence is not used in practice for large calculations. It occurs in some decision problem descriptions in theoretical computer science (e.g. some P-complete problems), where it is used to "artificially" decrease the run-time or space requirements of a problem. For instance, the problem of integer factorization is suspected to require more than a polynomial function of the length of the input as run-time if the input is given in binary, but it only needs linear runtime if the input is presented in unary. However, this is potentially misleading. Using a unary input is slower for any given number, not faster; the distinction is that a binary (or larger base) input is proportional to the base 2 (or larger base) logarithm of the number while unary input is proportional to the number itself. Therefore, while the run-time and space requirement in unary looks better as function of the input size, it does not represent a more efficient solution. In computational complexity theory, unary numbering is used to distinguish strongly NP-complete problems from problems that are NP-complete but not strongly NP-complete. A problem in which the input includes some numerical parameters is strongly NP-complete if it remains NP-complete even when the size of the input is made artificially larger by representing the parameters in unary. For such a problem, there exist hard instances for which all parameter values are at most polynomially large. Applications [edit] In addition to the application in tally marks, unary numbering is used as part of some data compression algorithms such as Golomb coding. It also forms the basis for the Peano axioms for formalizing arithmetic within mathematical logic. A form of unary notation called Church encoding is used to represent numbers within lambda calculus. Some email spam filters tag messages with a number of asterisks in an e-mail header such as X-Spam-Bar or X-SPAM-LEVEL. The larger the number, the more likely the email is considered spam. Using a unary representation instead of a decimal number lets the user search for messages with a given rating or higher. For example, searching for yield messages with a rating of at least 4. See also [edit] Unary coding One-hot encoding References [edit] ^ Hodges, Andrew (2009), One to Nine: The Inner Life of Numbers, Anchor Canada, p. 14, ISBN 9780385672665. ^ Davis, Martin; Sigal, Ron; Weyuker, Elaine J. (1994), Computability, Complexity, and Languages: Fundamentals of Theoretical Computer Science, Computer Science and Scientific Computing (2nd ed.), Academic Press, p. 117, ISBN 9780122063824. ^ Hext, Jan (1990), Programming Structures: Machines and Programs, vol. 1, Prentice Hall, p. 33, ISBN 9780724809400. ^ Brian Hayes (2001), "Third Base", American Scientist, 89 (6): 490, doi:10.1511/2001.40.3268, archived from the original on 2014-01-11, retrieved 2013-07-28 ^ Zdanowski, Konrad (2022), "On efficiency of notations for natural numbers", Theoretical Computer Science, 915: 1–10, doi:10.1016/j.tcs.2022.02.015, MR 4410388 ^ Woodruff, Charles E. (1909), "The Evolution of Modern Numerals from Ancient Tally Marks", American Mathematical Monthly, 16 (8–9): 125–33, doi:10.2307/2970818, JSTOR 2970818. ^ Hsieh, Hui-Kuang (1981), "Chinese Tally Mark", The American Statistician, 35 (3): 174, doi:10.2307/2683999, JSTOR 2683999 ^ Lunde, Ken; Miura, Daisuke (January 27, 2016), "Proposal to Encode Five Ideographic Tally Marks", Unicode Consortium (PDF), Proposal L2/16-046 ^ Sazonov, Vladimir Yu. (1995), "On feasible numbers", Logic and computational complexity (Indianapolis, IN, 1994), Lecture Notes in Comput. Sci., vol. 960, Springer, Berlin, pp. 30–51, doi:10.1007/3-540-60178-3_78, ISBN 978-3-540-60178-4, MR 1449655. See in particular p. 48. ^ Blaxell, David (1978), "Record linkage by bit pattern matching", in Hogben, David; Fife, Dennis W. (eds.), Computer Science and Statistics--Tenth Annual Symposium on the Interface, NBS Special Publication, vol. 503, U.S. Department of Commerce / National Bureau of Standards, pp. 146–156. ^ Hopcroft, John E.; Ullman, Jeffrey D. (1979), Introduction to Automata Theory, Languages, and Computation, Addison Wesley, Example 7.7, pp. 158–159, ISBN 978-0-201-02988-8. ^ Dewdney, A. K. (1989), The New Turing Omnibus: Sixty-Six Excursions in Computer Science, Computer Science Press, p. 209, ISBN 9780805071665. ^ Rendell, Paul (2015), "5.3 Larger Example TM: Unary Multiplication", Turing Machine Universality of the Game of Life, Emergence, Complexity and Computation, vol. 18, Springer, pp. 83–86, ISBN 9783319198422. ^ Arora, Sanjeev; Barak, Boaz (2007), "The computational model —and why it doesn't matter" (PDF), Computational Complexity: A Modern Approach (January 2007 draft ed.), Cambridge University Press, §17, pp. 32–33, retrieved May 10, 2017. ^ Moore, Cristopher; Mertens, Stephan (2011), The Nature of Computation, Oxford University Press, p. 29, ISBN 9780199233212. ^ Garey, M. R.; Johnson, D. S. (1978), "'Strong' NP-completeness results: Motivation, examples, and implications", Journal of the ACM, 25 (3): 499–508, doi:10.1145/322077.322090, MR 0478747, S2CID 18371269. ^ Golomb, S.W. (1966), "Run-length encodings", IEEE Transactions on Information Theory, IT-12 (3): 399–401, doi:10.1109/TIT.1966.1053907. ^ Magaud, Nicolas; Bertot, Yves (2002), "Changing data structures in type theory: a study of natural numbers", Types for proofs and programs (Durham, 2000), Lecture Notes in Comput. Sci., vol. 2277, Springer, Berlin, pp. 181–196, doi:10.1007/3-540-45842-5_12, ISBN 978-3-540-43287-6, MR 2044538. ^ Jansen, Jan Martin (2013), "Programming in the λ-calculus: from Church to Scott and back", The Beauty of Functional Code, Lecture Notes in Computer Science, vol. 8106, Springer-Verlag, pp. 168–180, doi:10.1007/978-3-642-40355-2_12, ISBN 978-3-642-40354-5. ^ Email, Spam Control, How to get service for departmental email servers External links [edit] OEIS sequence A000042 (Unary representation of natural numbers) Retrieved from " Categories: Numeral systems 1 (number) Elementary mathematics Coding theory Formal languages Hidden categories: CS1: long volume value Articles with short description Short description is different from Wikidata Pages using sidebar with the child parameter Commons link is locally defined
7468
https://people.math.harvard.edu/~knill/teaching/math1a_2011/handouts/math1a_2011.pdf
Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 1: What is Calculus? Calculus is a powerful tool to describe our world. It formalizes the process of taking differences and taking sums. Both are natural operations. Differences measure change, sums measure how things accumulate. We are interested for example in the total amount of precipitation in Boston over a year but we are also interested in how the temperature does change over time. The process of taking differences is in a limit called derivative. The process of taking sums is in the limit called integral. These two processes are related in an intimate way. In this first lecture, we want to look at these two processes in a discrete setup first, where functions are evaluated only on integers. We will call the process of taking differences a derivative and the process of taking sums as integral. Start with the sequence of integers 1, 2, 3, 4, ... . We say f(1) = 1, f(2) = 2, f(3) = 3 etc and call f a function. It assigns to a number a number. It assigns for example to the number 100 the result f(100) = 100. Now we add these numbers up. The sum of the first n numbers is called Sf(n) = f(1) + f(2) + f(3) + ... + f(n) . In our case we get 1, 3, 6, 10, 15, ... It defines a new function g which satisfies g(1) = 1, g(2) = 3, g(2) = 6 etc. The new numbers are known as the triangular numbers. From the function g we can get f back by taking difference: Dg(n) = g(n) −g(n −1) = f(n) . For example Dg(5) = g(5) −g(4) = 15 −10 = 5 and this is indeed f(5). Finding a formula for the sum Sf is not so easy. The young mathematician Karl-Friedrich Gauss realized as a 7 year old kid when giving the task to sum up the first 100 numbers that it is the same as adding up 50 times 101 which is 5050. Gauss found g(n) = n(n + 1)/2 . He did that by pairing things up. To add up 1 + 2 + 3 + . . . + 10 for example we can write this as (1 + 10) + (2 + 9) + (3 + 8) + (4 + 7) + (5 + 6) leading to n/2 terms of n + 1 if n is even. Taking differences again is easier Dg(n) = (n + 1)n/2 −n(n −1)/2 = n = f(n). Lets add up the new sequence again and compute h = Sg. We get the sequence 1, 4, 10, 20, 35, ... These numbers are called the tetrahedral numbers because one use h(n) marbles to build a tetrahedron of side length n. For example, we need h(4) = 20 golf balls for example to build a tetrahedron of side length 4. The formula which holds for h is h(n) = n(n + 1)(n + 2)/6 . We see that summing the differences gives the function in the same way as differencing the sum: SDf(n) = f(n) −f(0), DSf(n) = f(n) Don’t worry yet, if this is too abstract. We will come back to it again and again. But this is an arithmetic version of the fundamental theorem of calculus which we will explore in this course. The process of adding up numbers will lead to the integral R x 0 f(x) dx . The process of taking differences will lead to the derivative d dxf(x) . One of the high lights of this course is to understand the fundamental theorem of calculus: R x 0 d dtf(t) dt = f(x) −f(0), d dx R x 0 f(t) dt = f(x) and see why it is such a fantastic result. You see formally that it fits the result for difference and sum. A major goal of this course will be to understand the fundamental theorem result and see its use. But we have packed the essence of the theorem in the above version with S and D. It is a version which will lead us. 1 Problem: Given the sequence 1, 1, 2, 3, 5, 8, 13, 21, . . . which satisfies the rule f(x) = f(x − 1) + f(x −2). It defines a function on the positive integers. For example, f(6) = 8. What is the function g = Df, if we assume f(0) = 0? Solution: We take the difference between successive numbers and get the sequence of numbers 1, 0, 1, 1, 2, 3, 5, 8, ... After 2 entries, the same sequence appears again. We can also deduce directly from the above recursion that f has the property that Df(x) = f(x −2) . It is called the Fibonnacci sequence, a sequence of great fame. 2 Problem: Take the same function f given by the sequence 1, 1, 2, 3, 5, 8, 13, 21, ... but now compute the function h(n) = Sf(n) obtained by summing the first n numbers up. It gives the sequence 1, 2, 4, 7, 12, 20, 33, .... What sequence is that? Solution: Because Df(x) = f(x −2) we have f(x) −f(0) = SDf(x) = Sf(x −2) so that Sf(x) = f(x + 2) −f(2). Summing the Fibonnacci sequence produces the Fibonnacci sequence shifted to the left with f(2) = 1 is subtracted. It has been relatively easy to find the sum, because we knew what the difference operation did. This example shows: We can study differences to understand sums. The next problem illustrates this too: 3 Problem: Find the next term in the sequence 2 6 12 20 30 42 56 72 90 110 132 . Solution: Take differences 2 6 12 20 30 42 56 72 90 110 132 2 4 6 8 10 12 14 16 18 20 22 2 2 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0 . Now we can add an additional number, starting from the bottom and working us up. 2 6 12 20 30 42 56 72 90 110 132 156 2 4 6 8 10 12 14 16 18 20 22 24 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0 0 In the rest of this hour, we talk about some applied and not so applied problems which involve calculus. Homework 1 We have defined Sf(n) = f(1) + f(2) + ... + f(n) and Df(n) = f(n) −f(n −1) and seen f(n) = 1 we have g(n) = Sf(n) = n . f(n) = n we have g(n) = Sf(n) = n(n + 1)/2. f(n) = n(n + 1)/2 we have g(n) = Sf(n) = n(n + 1)(n + 2)/6. Guess a formula g(n) = Sf(n) forf(n) = n(n + 1)(n + 2)/6 and verify using algebraic manipulation that it satisfies Dg(n) = f(n). Can you see a pattern? 2 Find the next term in the sequence 3, 12, 33, 72, 135, 228, 357, 528, 747, 1020, 1353.... To do so, compute successive derivatives g = Df of f, then h = Dg until you see a pattern. 3 The function f(x) = 2x can first be defined on integers, then on rational numbers like 2(3/2) = √ 23. We have for example f(0) = 1, f(1) = 2, f(2) = 4, , . . . a) Verify that f satisfies the equation Df(x) = f(x −1), where Df(x) = f(x) −f(x −1). b) The function f(x) = 5x satisfies a similar rule. Which one? 4 Find g(n) = Sf(n) for the function f(n) = n2. This means we want to find a formula such that g(1) = 1, g(2) = 5, g(3) = 14 leading to the sequence of numbers 1, 5, 14, 30, 55, 91, 140, 204, 285, .. Note that we have already have computed Sf for g(n) = n(n + 1)/2 as well as for h(n) = n. Try to write f as a combination of g and h and use the rule D(f + g) = Df + Dg. 5 Find a formula g(n) = Sf(n) for the function f(n) = 7n. First compute the ”derivative” Df of f and go from there. General remarks about homework • Make sure to think about the problem yourself first before discussing it with others. • The time you spend on homework is valuable. Especially the exploration time before you know how to solve it. • If you do not know how to get started, don’t hesitate to ask. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 1: Worksheet In this first lecture, we want to see that the essence of calculus is already in basic arithmetic. Triangular numbers We stack disks onto each other building n layers and count the num-ber of discs. The number sequence we get are called triangular numbers. 1 3 6 10 15 21 36 45 ... n=1 n=2 n=3 n=4 This sequence defines a function on the natural numbers. For ex-ample, f(4) = 10. 1 Can you find f(100)? The task to find this number was given to Carl Friedrich Gauss in elementary school. The 7 year old came up quickly with an answer. How? Carl-Friedrich Gauss, 1777-1855 Tetrahedral numbers We stack spheres onto each other building n layers and count the number of spheres. The number sequence we get are called tetrahedral numbers. 1 4 10 20 35 56 84 120 ... Also this sequence defines a function. For example, g(3) = 10. But what is g(100)? Can we find a formula for g(n)? n=1 n=2 n=3 n=4 2 Once you know the formula for g(n) given to you as g(n) = n(n + 1)(n + 2)/6, verify that it is the right one, by checking g(n) −g(n −1) = n(n + 1)/2. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 2: Functions A function is a rule which assigns to a real number a new real number. An example is f(x) = x2−x. For example, it assigns to the number x = 3 the value 32−3 = 6. A function is given with a domain A, the points where f is defined and a codomain B a set of numbers in which f takes values. Typically, the codomain agrees with the set of real numbers and the domain to be all the numbers, where the function is defined. The function f(x) = 1/x for example is not defined at x = 0 so that we chose the domain A = R \ {0}, all numbers except 0. The function f(x) = 1/x takes values in the codomain R. If we choose A = B, then f(x) = 1/x reaches every point in B and is invertible. It is its own inverse. Here are a few examples of functions. We will look at them in more detail during the lecture, especially the polynomials, trigonometric functions and exponential function. identity f(x) = x constant f(x) = 1 linear f(x) = 3x + 1 quadratic f(x) = x2 cosine f(x) = cos(x) sine f(x) = sin(x) exponentials f(x) = exph(x) = (1 + h)x/h logarithms f(x) = logh(x) = exp−1 h power f(x) = 2x exponential f(x) = ex = exp(x) logarithm f(x) = log(x) = exp−1(x) absolute value f(x) = |x| devil comb f(x) = sin(1/x) bell function f(x) = e−x2 witch of Agnesi f(x) = 1 1+x2 sinc sin(x)/x We can build new functions by: add functions f(x) + g(x) scale functions 2f(x) translate f(x + 1) compose f(g(x)) invert f −1(x) difference f(x + 1) −f(x) sum up f(x) + f(x + 1) + . . . Here are important functions: polynomials x2 + 3x + 5 rational functions (x + 1)/(x4 + 1) exponential ex logarithm log(x) trig functions sin(x), tan(x) inverse trig functions arcsin−1(x), arctan(x). roots √x, x1/3 We will look at these functions a lot during this course. The logarithm, exponential and trigono-metric functions are especially important. For some functions, we need to restrict the domain, where the function is defined. For the square root function √x or the logarithm log(x) for example, we have to assume that the number is positive. We write that the domain is (0, ∞) = R+. For the function f(x) = 1/x, we have to assume that x is different from zero. Keep these three examples in mind. The graph of a function is the set of points {(x, y) = (x, f(x)) } in the plane, where x runs over the domain A of f. Graphs allow us to visualize functions. We can ”see them”, when we draw the graph. expHxL x logHxL x e-x2 x x sinH1xL x x x x3 - 3 x x Homework 1 Draw the function f(x) = x + sin(x). Its graph goes through the origin (0, 0). a) A function is called odd if f(−x) = −f(x). Is f odd? b) A function is called even if f(x) = f(−x). Is f even? c) A function is called monotone increasing if f(y) > f(x) if y > x. Is f monotone increasing? You do not have to decide this yet analytically. Just draw(∗) the function and make up your mind. 2 A function f : A →B is called invertible or one to one if there is an other function g such that g(f(x)) = x for all x in A and f(g(y)) = y for all y ∈B. For example, the function g(x) = √x is the inverse of f(x) = x2 as a function from A = [0, ∞) to B = [0, ∞). Determine from the following functions whether they are invertible. If they are invertible, find the inverse. a) f(x) = sin(x) from A = [0, π/2] to B = [0, 1] b) f(x) = x3 from A = R to B = R c) f(x) = x6 from A = R to B = R d) f(x) = exp(5x) from A = R to B = R+ = (0, ∞). e) f(x) = 1/(1 + x2) from A = [0, ∞) to B = [0, ∞). 3 Look at the function f1(x) = sin(x), f2(x) = sin(sin(x)), f3(x) = sin(sin(sin(x))). a) Draw the graphs of the functions f1, f2, f3 on the interval [0, 4π]. b) Can you imagine what f100000(x) looks like? You might want to make more experiments here to see the answer. Of course you are allowed to plot the functions with a calculator or with an online grapher like Wolfram alpha. (The weblink can be found below). 4 Lets call a function f(x) a composition square root of a function g if f(f(x)) = g(x). For example, the function f(x) = x2 + 1 is the composition square root of g(x) = x4 + 2x2 + 2 because f(f(x)) = (x2 + 1)2 + 1 = g(x). Find the composition square roots of the following functions: a) f(x) = sin(sin(x)). b) f(x) = x4 c) f(x) = x d) f(x) = x4 + 2x2 + 2 e) f(x) = eex. Note that it can be difficult in general to find the square root function in general. Already for basic functions like exp(x) or sin(x), we are speechless. 5 A function f(x) has a root at x = a if f(a) = 0. Roots are places, where the function is zero. Find one root for each of the following functions or state that there is none. a) f(x) = sin(x) b) f(x) = exp(x) c) f(x) = x3 −x d) f(x) = sin(x)/x −1 e) f(x) = csc(x) = 1/ sin(x) () Here is how you can use the Web to plot a function. The example given is sin(x). ✞ http ://www. wolframalpha . com/ input /? i=Plot+sin (x) ✝ ✆ Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 2: Worksheet In this lecture, we want to learn what a function is and get acquainted with the most important examples. Trigonometric functions The cosine and sine functions can be defined geometrically by the co-ordinates (cos(x), sin(x)) of a point on the unit circle. The tangent function is defined as tan(x) = sin(x)/ cos(x)). cos(x) = adjacent side/hypothenuse sin(x) = opposite side/hypothenuse tan(x) = opposite side/adjadcent side Pythagoras theorem gives us the important identity cos2(x) + sin2(x) = 1 Define also cot(x) = 1/ tan(x). Less important but sometimes used are sec(x) = 1/ cos(x), csc(x) = 1/ sin(x). 1 Find cos(π/3), sin(π/3). 2 Where does cos and sin have roots, places, where the function is zero? 3 Find tan(3π/2) and cot(3π/2). 4 Find cos(3π/2) and sin(3π/2). 5 Find tan(π/4) and cot(π/4). cosHΦL sinHΦL Φ 1 cosHxL x 2Π sinHxL x 2Π tanHxL x А2 -А2 The exponential function The function 2x is first of all defined for all integers like 210 = 1024. By taking roots, we can define it for rational numbers like 23/2 = 81/2 = √ 8 = 2.828.... Since the function 2x is monotonone on the set of rationals, we can fill the gaps and define the function 2x for any x. By taking square roots again and again, we see 21/2, 21/4, 21/8, ... we approach 20 = 1. 2x x There is nothing special about 2 and we can take any positive base a and define the exponential ax. It satisfies a0 = 1 and the remarkable rule: ax+y = ax · ay It is spectacular because it provides a link between addition and mul-tiplication. We will especially consider the exponential exph(x) = (1 + h)x/h , where h is a positive parameter. This is a supercool exponential be-cause it satisfies exph(x + h) = (1 + h) exph(x) so that [exph(x + h) −exph(x)]/h = exph(x) . Hold on to that. We will look at this later again. In modern language, we would say that ”the quantum derivative of the quantum exponen-tial is the function itself for any Planck constant h”. For h = 1, we have the function 2x we have started with. In the limit h →0, we get the important exponential function exp(x) which we also call ex. For x = 1, we get the Euler number e = e1 = 2.71828.... 1 What is 2−5? 2 Find 21/2. 3 Find 271/3. 4 Why is A = 23/4 smaller than B = 24/5? Take the 20th power of both numbers. 5 Assume h = 2 find exph(4). Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 3: Limits Sometimes, functions look as if they are not defined at some point. They often allow a continuation to ”non-allowed” places however. Lets look at some examples: 1 The function f(x) = (x3 −1)/(x −1) is at first not defined at x = 1. However, for x close to 1, nothing really bad happens. We can evaluate the function at points closer and closer to 1 and get closer and closer to 3. We will say limx→1 f(x) = 3. Indeed, as you might have seen already, we have f(x) = x2 + x + 1 by factoring out the term x −1. While the function was initially not defined at x = 1, we can assign a natural value 3 at the point x = 1 and keep a ”nice” function. The graph will continue nicely through that point. Definition. We write x →a if we mean that the number x approaches a from either side. A function f(x) has a limit at a point a if there exists b such that f(x) →b for x →a. We write limx→a f(x) = b. It should not matter, whether we approach a from the left or from the right. In both cases, we should get the same limiting value b. 2 The function f(x) = sin(x)/x is called sinc(x). It converges to 1 as x →0. We can see this geometrically by comparing the side a = sin(x) of a right angle triangle with a small angle α = x and hypotenuse 1 with the length of the arc between B, C of the unit circle centered at A. The arc has length x which is close to sin(x) for small x. Keep this example in mind. It is a good one. Remark. It is possible to see this analytically. A computer for example approximates the function sin(x) with the polynomial x −x3/3! + x5/5! −· · · + x100/100! and if we divide this by x, we get 1 −x2/3! + x4/5! −· · · + x99/100! which converges to 1 as x approaches 0. 3 The quadratic function f(x) has the property that f(x) approaches 4 if x approaches 2. This is a very typical case. To evaluate functions at a point, we do not have to take a limit. The function is already defined there. This is important: given a typical function, most points are ”healthy”. We do not have to worry about limits there. In most cases we see in real pplications we only have to worry about limits when the function divides by 0. For example f(x) = (x4 + x2 + 1)/x needs to be investigated carefully only at x = 0. You see for example that for x = 1/1000, the function is slightly larger than 1000, for x = 1/1000000 it is larger than one million. There is no rescue here. The limit does not exist at 0. 4 More generally, for all polynomials, the limit limx→a f(x) = f(a) is defined. We do not have to worry about limits, if we deal with polynomials. 5 For all trigonometric polynomials involving sin and cos, the limit limx→a f(x) = f(a) is defined. We do not have to worry about limits if we deal with trigonometric polynomials like sin(3x) + cos(5x). The function tan(x) however has no limit at x = π/2. There is no value b we can find so that tan(π/2 + h) →b for h →0. This is due to the fact that cos(x) is zero at π/2. We have tan(x) goes to +∞”plus infinity” for x ց π/2 and tan(x) goes to −∞for x ր π/2. In the first case, we approach π/2 from the right and in the second case from the left. 6 The cube root function f(x) = x1/3 converges to 0 as x →0. For x = 1/1000 for example, we have f(x) = 1/10 for x = 1/n3 the value f(x) is 1/n. The cube root function is defined everywhere on the real line, like f(−8) = −2 and is continuous everywhere. 1 2 Why do we worry about limits at all? One of the main reasons will is that we will define the derivative and integral using limits. But we will also use limits to get numbers like π = 3.1415926, ..... In the next lecture, we will look at the important concept of continuity, which involves limits too. a b x fHxL Figure: We can test whether a function has the limit b at a point a if for every vertical interval I containing b there exists a horizontal interval J containing a such that if x is in J, then f(x) is in I. If the function stays bounded, does not oscillate at the point like sin(1/x) or jump, then the limit exists. x Figure: We see here the function f(x) = arctan(tan(x) + 1), where arctan is the inverse of tan giving the angle from the slope. In this case, the limit does not exist for a = π/2. If we approach this point a from the right, we are always far below the limiting value. The limit exists from the left if we postulate f(π/2) = π/2. Note that f has a priori no value at x = π/2 because tan(x) becomes infinite there. 3 7 Problem: Determine from the following functions whether the limits limx→0 f(x) exist. If the limit exists, find it. a) f(x) = cos(x)/ cos(2x) b) f(x) = tan(x)/x c) f(x) = (x2 −x)/(x −1) d) f(x) = (x4 −1)/(x2 −1) e) f(x) = (x + 1)/(x −1) f) f(x) = x/ sin(x) g) f(x) = sin(x)/x2 h) f(x) = sin(x)/ sin(2x) Solution: a) There is no problem at x = 0. Both, the nominator and denominator converge to 1. The limit is 1 b) This is sinc(x)/ cos(x). There is no problem at x = 0 for sinc nor for 1/ cos(x). The limit is 1 . c) We can heal this function. It is the same as x + 1. The limit is 1 . d) We can heal this function. It is the same as x2 + 1. The limit is 2 . e) There is no problem at x = 0. There is mischief at x = 1 although but that is far, far away. At x = 0, we get 1 . f) This is the prototype. We know that the limit is 1 . g) This limit does not exist. Because it is sinc(x)/x. Because sinc(x) converges to 1. we are in trouble when dividing again by x. There is no limit. h) We know sin(x)/x →1 so that also sin(2x)/(2x) has the limit 1. If we divide them, see sin(x)/ sin(2x) →1/2. The result is 1/2 . 4 Homework 1 a) Draw the graph of the function f(x) = (1 −cos(x)) x2 . b) Where is the function f defined? Can you find the limit at the places, where it is not defined? c) A function is even if f(x) = f(−x), odd if f(x) = −f(x). Is f even or odd, or neither? d) What happens with the function f in the limit x →+∞and x →−∞? 2 Find the limits of each of the following functions at the point x →0: a) f(x) = (x4 −1)/(x −1) b) f(x) = sin(3x)/x c) f(x) = sin(5x)/x d) f(x) = sin(3x)/ sin(5x) 3 a) Can you see the limit of g(h) = [f(x + h) −f(x)]/h as a function of h at the point x = 0 for the function f(x) = sin(x)? b) Verify that the function f(x) = exph(x) = (1+h)x/h satisfies [f(x+h)−f(x)]/h = f(x). We define ex = exp(x) = limh→0 exph(x). 4 Find the limits for x →0: a) f(x) = (x2 −2x + 1)/(x −1). b) f(x) = 2x. c) f(x) = 22x. d) f(x) = sin(sin(x))/ sin(x). 5 We explore in this problem the limit of the function f(x) = xx if x →0. Can we find a limit? Take a calculator or use Wolfram α and experiment. What do you see when x →0? Optional: can you find a explanation for your experiments? Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 3: Worksheet We study a few limits. The Sinc function A prototype function for studying limits is the sinc function f(x) = sin(x) x . It is an important function and appears in many applications like in the study of waves or signal processing (it is used in low pass filters). The name sinc comes from its original latin name sinus cardinalis. sincHxL x 2Π 1 Does the function cos(x) x have a limit at x →0? 2 Does the function sin(x2) x2 have a limit for x →0? 3 Does the function sin(x2) x have a limit for x →0? 4 Does the function sin2(x) x2 have a limit for x →0? 5 Does the limit 1−cos2(x) x2 exist for x →0? 6 Does the function x sin(x) have a limit for x →0? 7 Does the function sin(x) |x| have a limit for x →0? 8 Does the function sin(x) √ |x| have a limit for x →0? Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 4: Continuity A function f is called continuous at a point p if a value f(p) can be found such that f(x) →f(p) for x →p. A function f is called continuous on [a, b] if it is continuous for every point x in the interval [a, b]. In the interior (a, b), the limit needs to exist both from the right and from the left. At the bound-ary a only the right limit needs to exist and at b only the left limit. Intuitively, a function is continuous if you can draw the graph of the function without lifting the pencil. Continu-ity means that small changes in x results in small changes of f(x). 1 Any polynomial is continuous everywhere. To see this note that the sum of two continuous functions is continuous and that a multiple of a continuous function is continuous. Since xn is continuous for all n, and every polynomial is a sum of multiples of such functions, we have continuity in general. 2 The function f(x) = 1/x is continuous everywhere except at x = 0. It is a prototype of a function which is not continuous due to a pole. The source for the trouble is the division by zero which would happen if we would try to evaluate the function at x = 0. 3 The function csc(x) = 1/ sin(x) is not continuous at x = 0, x = π, x = 2π and any multiple of π. It has poles there because sin(x) is zero there and because we would divide by zero at such points. 4 The function f(x) = sin(π/x) is continuous everywhere except at x = 0. It is a prototype of a function which is not continuous due to oscillation. We can approach x = 0 in ways that f(xn) = 1 and such that f(zn) = −1. Just chose xn = 2/(4k + 1) and zn = 2/(4k −1). 5 The signum function f(x) = sign(x) =    1 x > 0 −1 x < 0 0 x = 0 is not continuous at 0. It is a prototype of a function which has a jump discontinuity at 0. We can refine the notion of continuity and say that a function is continuous from the right, if there exists a limit from the right limx↓a f(x) = b. Similarly a function f can be continuous from the left only. Most of the time we mean with ”continuous”= ”continuous on the real line”. Rules: a) If f and g are continuous, then f + g is continuous. b) If f and g are continuous, then f ∗g is continuous. c) If f and g are continuous and if g > 0 then f/g is continuous. d) If f and g are continuous, then f ◦g is continuous. 6 √ x2 + 1 is continuous everywhere on the real line. 7 cos(x) + sin(x) is continuous everywhere. 8 The function f(x) = log(|x|) is continuous everywhere except at 0. Indeed since for every integer n, we have f(e−n) = −n, this can become arbitrarily large for n →∞even so e−n converges to 0 for n running to infinity. 9 While log(|x|) is not continuous at x = 0, the function 1/ log |x| is continuous at x = 0. Is it continuous everywhere? 10 The function f(x) = [sin(x + h) −sin(x)]/h is continuous for every h > 0. We will see next week that nothing bad happens when h becomes smaller and smaller and that the continuity will not deteriorate. Indeed, we will see that we get closer and closer to the cos function. There are three major reasons, why a function is not continuous at a point: it can jump, oscillate or escape to infinity. Here are the prototype examples. We will look at more during the lecture. sign x x x 1x x Why do we like continuity? We will see many reasons during this course but for now lets just say that: A wild continuous function. This Weierstrass function is believed to be a fractal. ”Continuity tames a func-tion. It can be pretty wild, but not too crazy.” A crazy discontinuous function. It is discontinuous at every point and known to be a fractal. Continuity will be useful later for extremization. A continuous function on an interval [a, b] has a maximum and minimum. And if a continuous function is negative at some place and positive at an other, there is a point between, where it is zero. These are all useful properties to have and they do not hold if a function is not continuous. 11 Problem Determine from each of the following functions, where discontinuities appear and give a short reason. a) f(x) = log(|x2 −1|) b) f(x) = sin(cos(π/x)) c) f(x) = cot(x) + tan(x) + x4 d) f(x) = x4 + 5x2 −3x + 4 e) f(x) = x2−x x Solution. a) log(|x|) is continuous everywhere except at x = 0. Since x2 −1 = 0 for x = 1 or x = −1, the function f(x) is continuous everywhere except at x = 1 and x = −1. b) The function π/x is continuous everywhere except at x = 0. Therefore cos(cos(π/x) is continuous everywhere except possibly at x = 0. We have still to investigate the point x = 0 but there, the function cos(π/x) takes values between −1 and 1 for points arbitrarily close to x = 0. The function f(x) takes values between sin(−1) and sin(1) arbitrarily close to x = 0. It is not continuous there. c) The function x4 is continuous everywhere. We do not have to consider it. The function tan(x) is continuous everywhere except at the points points kπ, integer multiples of π. The function cot(x) is continuous everywhere except at points π/2 + kπ. The function f is therefore continuous everywhere except at the point x = kπ/2, multiples of π/2. d) The function is a polynomial. We know that polynomials are continuous everywhere. e) The function is continuous everywhere except at x = 0, where we have to look at the function more closely. But we can heal the function by dividing nominator and denominator by x which is possible for x different from 0. We get x −1. Homework 1 On which intervals is the following function continuous? -1 1 2 3 4 5 6 -4 -2 2 4 6 2 For the following functions, determine the points, where f is not continuous. a) f(x) = tan(1 −x) b) x cos(1/x) c) sign(x)/x d) sinc(x) + sin(x) + x8 + log(x) e) x2+5x+x4 x−1 State which kind of discontinuity appears. 3 Construct a function which has a jump discontinuity, an oscillatory one as well as an escape to infinity. Can you construct an example where two of these flaws happen at the same point? Can you even construct an example where all three happen at the same point? 4 Heal the following functions: a) (x5 −32)/(x −2) b) x5 −x3/(x2 −1) c) ((sin(x))3 −sin(x))/ sin(x). d) (x3 + 3x2 + 3x + 1)/(x2 + 2x + 1) e) (x1000 −1)/(x100 −1) 5 Is the following function continuous? cos(cos(cos(cos(cos(x)))) sin(sin(sin(e(e(e(e(e(e(e(e(e(e(eex))))))))))))) log(2x+1)+2+cos((x)) Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 4: Worksheet Whats good and whats bad? We have seen that oscillation, poles and jumps are the perils for continuity. In general, we do not have to worry about continuity. There are very few mechanisms which bring you in peril. A function can either start to oscillate like mad, rush to infinity or jump. All cases are usually due to division by zero somewhere. Good Guys Bad Guys x2 + 4x + 6 1/x at 0 sin(x), cos(x) tan(x) at π/2 exp(x) log |x| at 0 sinc(x) = sin(x) x sec(x) = 1 cos(x) at π/2 Which functions are continuous? Which of the following functions are continuous? 1 Is f(x) = r|x| continuous at x = 0? 2 Is f(x) = 1 √ |x| continuous at x = 0? 3 Is 1 log |x2| continuous at x = 0? 4 Is log(log |x|) continuous at x = 0? 5 Is 1/(1 + 1/(x4 + 1)) continuous everywhere? 6 Is sin(sec(x)) continuous everywhere? Enemy of continuity Oscillations, escape to infinity and jumps are reasons for discontinuity. x 1x x sign x x Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 5: Intermediate Value Theorem If f(a) = 0, then the value a is called a root of f. For example, f(x) = cos(x) has the root x = π/2. 1 f(x) = 4x + 6. Find the roots of f. Answer: set the function equal to 0 and solve for x. We get 4x + 6 = 0 2 f(x) = x2+2x+1 Find the roots of f. Answer: we can write f(x) = (x+1)2. The function has the root x = −1. 3 f(x) = (x −2)(x + 6)(x + 3). Find the roots of f. 4 f(x) = 12 + x −13x2 −x3 + x4. Find the roots of f. We do not have a formula for this, but we can try. Indeed, we see that for x = 1, x = −3, x = 4, x = −1 we have roots. 5 f(x) = exp(x). This function does not have any root. 6 f(x) = 2x −16 has the root x = 2. Intermediate value theorem of Bolzano. If f is continu-ous on [a, b] and f(a), f(b) have different signs, there is a root of f in (a, b). Proof. We can assume f(a) < 0 and f(b) > 0. The other case is similar. Look at the point c = (a + b)/2. If f(c) < 0, then look take [c, b] as your new interval, otherwise, take [a, c]. We get a new root problem on a smaller interval. Repeat the procedure. After n steps, the search is narrowed to an interval [un, vn] of size 2−n(b −a). Continuity assures that f(un) −f(vn) →0 and f(un), f(vn) have different signs. Both un, vn converge to a root of f. 7 The function f(x) = x17 −x3 + x5 + 5x7 + sin(x) has a root. Solution. The function goes to +∞for x →∞and to −∞for x →−∞. We have for example f(10000) > 0 and f(−1000000) < 0. The intermediate value theorem assures there is a point where f(x) = 0. 8 There is a solution to the equation xx = 10. Solution: for x = 1 we have xx = 1 for x = 10 we have xx = 1010 > 10. Apply the intermediate value theorem. 9 There exists a point on the earth, where the temperature is the same as the temperature on its antipode. Solution: Lets draw a meridian through the north and south pole and let f(x) be the temperature on that circle. Define g(x) = f(x) −f(x + π). If this function is zero on the north pole, we have found our point. If not, g(x) different signs on the north and south pole. There exists therefore a point, where the temperature is the same. 10 Wobbly Table Theorem. On an arbitrary floor, a square table can be turned so that it does not wobble any more. Why? The 4 legs ABCD are on a square. Let x be the angle of the line AC with with some coordinate axes if we look from above. Given the angle x, we can position the table uniquely as follows: the center of ABCD is on the z-axes, the legs ABC are on the floor and AC points in the direction x. Let f(x) denote the height of the fourth leg D from the ground. If we find an angle x such that f(x) = 0, we have a position where all four legs are on the ground. Assume f(0) is positive. (If it is negative, the argument is similar.) Tilt the table around the line AC so that the two legs B,D have the same vertical distance h from the ground. Now translate the table down by h. This does not change the angle x nor the center of the table. The two previously hovering legs BD now touch the ground and the two others AC are below. Now rotate around BD so that the third leg C is on the ground. The rotations and lowering procedures have not changed the location of the center of the table nor the direction. This position is the same as if we had turned the table by π/2. Therefore f(π/2) < 0. The intermediate value theorem assures that f has a root between 0 and π/2. Define Df(x) = (f(x+h)−f(x))/h. Lets call it the derivative of f for the constant h. We will study it more in the next lecture. But you have verified for example D exph(x) = exph(x) in a homework. Lets call a point p, where Df(x) = 0 a critical point for h. Lets call a point a a local maximum if f(a) ≥f(x) in an open interval containing a. Define similarly a local minimum as a point where f(a) ≤f(x). 11 The function f(x) = x(x −h)(x −2h) has the derivative Df(x) = 3x(x −h) as you have verified in the case h = −1 in the first lecture of this course in a worksheet. We will write [x]3 = x(x −h)(x −2h) and [x]2 = x(x −h). The computation just done tells that D[x]3 = 3[x]2. Since [x]2 has exactly two roots 0, h, the function [x]3 has exactly 2 critical points. 12 More generally for [x]n+1 = x(x −h)(x −2h)...(x −nh) we have D[x]n+1 = (n + 1)D[x]n. Because [x]n has exactly n roots, the function [x]n+1 has exactly n critical points. Keep the formula D[x]n = n[x]n−1 in mind! 13 The function exph(x) = (1 + h)x/h satisfies D exph(x) = exph(x). Because this function has no roots and the derivative is the function itself, the function has no critical points. Indeed, this function is monotone. Figure: We see the function [x]4 = x(x −h)(x −2h)(x −3h) with h = 0.5. This function has 3 critical points because D[x]4 = 4[x]3 and [x]3 has roots at 0, h, 2h. There are three local maxima or minima according to the theorem. Later in the course, we will look at the derivative Df in the limit when h →0. And then the critical points are places where the tangent is horizontal. In our case now, a critical point is a point so that if we walk by a step h to the right, the function does not change. For now, just remember the formula D[x]n = n[x]n−1. It will be the same formula later on when we go to the limit h →0. Critical points lead to extrema as we will see later in the course. In our discrete setting we can say: Fermat’s maximum theorem If f is continuous and has a critical point a for h, then f has either a local maximum or local minimum inside the open interval (a, a + h). Look at the range of the function f restricted to [a, a + h]. It is a bounded interval [c, d] by the intermediate value theorem. There exists especially a point u for which f(u) = c and a point v for which f(v) = d. These points are different if f is not constant on [a, a + h]. There is therefore one point, where the value is different than f(a). If it is larger, we have a local maximum. If it is smaller we have a local minimum. 14 Problem. Verify that a cubic polynomial has maximally 2 critical points. Solution f(x) = ax3 + bx2 + cx + d. Because the x3 terms cancel in f(x + h) −f(x), this is a quadratic polynomial. It has maximally 2 roots. Homework 1 Find the roots for f(x) = −30 + 49x −19x2 −x3 + x4 2 Use the intermediate value theorem to find a root of f(x) = x2 −6x + 8 on [0, 3]. Are all roots in this interval? 3 a) Argue why there was a time, when Lady Gaga’s height was exactly 1 meter and not one mm more less. b) And that there was a time, when she weighed 50 kg and not a milligram more or less. c) Was there a time, when she owned exactly 1’000’000 dollars and not one dime more or less? 4 Argue why there is a solution to a) cos(x) = x. b) exp(x) = x. c) sinc(x) = x4. 5 a) Draw the graph of f(x) = x3 −x. b) Locate the local maxima and minima. c) Find the critical points of f to the constant h = 1. That means, find the places, where f(x + 1) −f(x) = 0. d) For every point a you have found in c), verify that there is a local maximum or minimum in [a, a + 1]. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 5: Worksheet Its groundhog day and a blizzard is coming. We study extrema and the intermediate value theorem. The intermediate value theorem 1 Today on groundhog day, the average temperature is 33◦Fahren-heit. Last summer, there was an average temperature was 77.2◦. Was there a time between July 1, 2010 and Feb 2, 2011, when the temperature was exactly 50◦? 2 We have got 38 inches of snow this month already. Does this mean there was a time that we had 20 inches of snow on the ground? 3 Is there a point x, where 1/ sin(x) = 1/2? Why does the intermediate value theorem not give such a point? We have 1/ sin(π/2) = 1 and 1/ sin(3π/2) = −1. 4 Is there a point, where sign(x) = 1/2? Remember the signum function. It is 1 for positive numbers, 0 for 0 and −1 for negative numbers. 5 Lets call the function f(x) = x −floor(x) the ground hog function. If you know the movie with Bill Murray, you know why. Can you find an interval on which the intermediate value theorem fails? Feb 2 Feb 2 Feb 2 0 1 2 The derivative and extrema 6 Find a concrete function which has only one local maximum, and no local minimum. 7 We have seen a remarkable theorem assuring the existence of maxima and minima. In the classical sense this is not true. We will define critical points as points, where f ′(x) = 0 and see that for f(x) = x3, the derivative is 3x2 which is zero at x = 0. Does f(x) have a local maximum or minimum at x = 0? Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 6: Some examples Here are some worked out examples, similar to what we expect you to do for the homework of lecture 6: The homework should be straightforward, except when finding Sf(x), we want to add a constant such that Sf(0) = 0. In general, you will not need to evaluate functions and can leave terms like sin(5x) as they are. If you have seen calculus already, then you could do this exercice by writing d dxf(x) instead of Df(x) and by writing Z x 0 f(x) dx instead of Sf(x). We did not introduce the derivative d f/dx nor the integral R x 0 yet. For now, just use the Differentiation rules and integrations rules in the box to the right to solve the problem. 1 Problem: Find the derivative Df(x) of the function f(x) = sin(5 · x) + x7 + 3. Answer: From the differentiation rules, we know Df(x) = 5 cos(5 · x) + 7x6 . 2 Problem: Find the derivative Df(0) of the same function f(x) = sin(5 · x) + 5x7 + 3. Answer: We know Df(x) = 5 cos(5 · x) + 35x6. Plugging in x = 0 gives 5 . 3 Problem: Find the integral Sf(x) of the function f(x) = sin(5 · x) + 5x7 + 3. Answer: From the integration rules, we know Sf(x) = −cos(5 · x)/5 + 5x8/8 + 3x . 4 Problem: Find the integral Sf(1) of the function f(x) = x2 + 1. Answer: From the integration rules, we know Sf(x) = x3/3 + x. Plugging in x = 1 gives 1/3 + 1 if we use the functions in the limit h →0. For positive h, we have to evaluate x(x −h)(x −2h)/3 + x for x = 1 which is (1 −h)(1 −2h)/3 + 1 5 Problem: Find the integral Sf(1) of the function f(x) = exp(4 · x). Answer: From the integration rules, we know Sf(x) = exp(4 · x)/4 −1/4. We have added a constant such that Sf(0) = 0. Plugging in x = 1 gives exp(4)/4 −1/4 . 6 Problem: Assume h = 1/1000. Determine the value of 1 1000[f( 0 1000) + f( 1 1000) + ... + f( 999 1000)] for the function f(x) = −sin(7x) + exp(3x). Answer: The problem asks for Sf(1). We first compute Sf(x) taking care that Sf(0) = 0. Sf(x) = cos(7x)/7 + exp(3x)/3 −(1/7 + 1/3) . Now plug in x = 1 to get cos(7)/7 + exp(3)/3 −(1/7 + 1/3) . Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 6: Fundamental theorem Calculus is the theory of differentiation and integration. We fix here a positive constant h and take differences and sums. Without taking limits, we prove a version of the fundamental theorem of calculus and differentiate and integrate polynomials, exponentials and trigonometric functions. Given a function, define the differential quotient Df(x) = (f(x + h) −f(x)) 1 h If f is continuous then Df is a continuous function too. We call it also ”derivative”. 1 Lets take the constant function f(x) = 5. We get Df(x) = (f(x+h)−f(x))/h = (5−5)/h = 0 everywhere. We see that in general if f is a constant function, then Df(x) = 0. 2 f(x) = 3x. We have Df(x) = (f(x + h) −f(x))/h = (3(x + h) −3x)/h which is 3 . 3 If f(x) = ax + b, then Df(x) = a . For constant functions, the derivative is zero. For linear functions, it is the slope. 4 For f(x) = x2 we compute Df(x) = ((x + h)2 −x2)/h = (2hx + h2)/h which is 2x + h . Given a function f define a new function Sf(x) by summing up all values of f(hj) where 0 ≤jh < x. That is, if k is such that kh is the largest below x, then Sf(x) = h[ f(0) + f(h) + f(2h) + .... + f(kh) ] We call Sf also the ”integral” or ”antiderivative” of f. 5 Compute Sf(x) for f(x) = 1. Solution. We have Sf(x) = 0 for x ≤h, and Sf(x) = h for h ≤x < 2h and Sf(x) = 2h for 2h ≤x < 3h. In general S1(jh) = j and S1(x) = kh where k is the largest integer such that kh < x. The function g grows linearly but quantized steps. The difference Df(x) will become the derivative f ′(x) . The sum Sf(x) will become the integral R x 0 f(t) dt . Df means rise over run and is close to the slope of the graph of f. Sf means areas of rectangles and is close to the area under the graph of f. k h x y 0 h fHk hL-fH0L k h x y 0 h fHk hL Theorem: Sum the differences and get SDf(kh) = f(kh) −f(0) Theorem: Difference the sum and get DSf(kh) = f(kh) 6 For f(x) = [x]m h = x(x −h)(x −2h)...(x −mh + h) we have f(x + h) −f(x) = (x(x −h)(x −2h)...(x −kh + 2h)) ((x + h) −(x −mh + h)) = [x]m−1hm and so D[x]m h = mx h . Lets leave the h away to get the important formula D[x]m = m[x]m−1 We can establish from this differentiation formulas for polynomials. 7 If f(x) = [x] + [x]3 + 3[x]5 then Df(x) = 1 + 3[x]2 + 15[x]4. The fundamental theorem allows us to integrate and get the right values at the points k/n: 8 Find Sf for the same function. The answer is Sf(x) = [x]2/2 + [x]4/4 + 3[x]6/6. Define exph(x) = (1+h)x/h. It is equal to 2x for h = 1 and morphs into the function ex when h goes to zero. As a rescaled exponential, it is continuous and monotone. 9 The function exph(x) = (1 + h)x/h satisfies D exph(x) = exph(x). Solution: exph(x + h) = (1 + h) exp(x) shows that. D exph(x) = exph(x) 10 Define expa(x) = (1 + ah)x/h. Now D expa h(x) = a expa h(x). Since exph(ax) is not equal to expa h(x), we write also ea·x h = exph(a · x) = expa h(x). Now: D exph(a · x) = a exph(a · x) 11 We can also replace a with the complex ai and consider expai h (x) = (1 + aih)x/h. Now, D expai h (x) = ai expai h (x). Real and imaginary parts define new functions expai h (x) = cosh(a · x) + i sinh(a · x). We have D sinh(a · x) = a cosh(a · x) and D cosh(a · x) = −a sinh(a · x). These functions morph into the familiar cos and sin functions for h →0. But in general, for any h and any a, we have D cosh(a · x) = −a sinh(a · x) and D sinh(a · x) = a cosh(a · x). Homework We leave the h away in this homework. To have more fun, also define logh as the inverse of exph and define 1/[x]h = D logh(x) for x > 0. If we start integrating from 1 instead of 0 as usual we have S11/[x]h = logh(x). 1 We also write here xn for [x]n h and write exp(a · x) = ea·x instead of expa h(x) and log(x) instead of logh(x) because we are among friends. Use the differentiation and integration rules on the right to find derivatives and integrals of the following functions: 1 Find the derivatives Df(x) of the following functions: a) f(x) = x2 + 6x7 + x a) f(x) = x4 + log(x) c) f(x) = −3x3 + 17x2 −5x. What is Df(0)? 2 Find the integrals Sf(x) of the following functions: a) f(x) = x4. b) f(x) = x2 + 6x7 + x c) f(x) = −3x3 + 17x2 −5x. What is Sf(1)? 3 Find the derivatives Df(x) of the following functions a) f(x) = exp(3 · x) + x6 b) f(x) = 4 exp(−3 · x) + 9x6 c) f(x) = −exp(5 · x) + x6 4 Find the integrals Sf(x) of the following functions a) f(x) = exp(6 · x) −3x6 b) f(x) = exp(8 · x) + x6 c) f(x) = −exp(5 · x) + x6 5 Define f(x) = sin(4 · x) −exp(2 · x) + x4 and assume h = 1/100 in part c). a) Find Df(x) b) Find Sf(x) c) Determine the value of 1 100[f( 0 100) + f( 1 100) + · · · + f( 99 100)] . 1We do not see h in daily lives, or do we? An allegory: in our universe, where h = 1.616 · 10−35m, the difference between the sinh and sin is so small that a x-ray oscillating with ν = 1017 Herz traveling for 13 billion years t = 4 · 1017s would only start to deviate noticeably from the classical sin(x) wave when it reaches us at ν · t = 4 · 1034 oscillations. Since sinh(x) −sin(x) only starts to grow at around x = 1/h ∼1035 oscillations, the x-ray would look the same when using the trig functions sinh, cosh. If γ is in the Gamma ray spectrum 1019Hz, the functions sinh, cosh start to grow in amplitude earlier. A γ wave emitted 1 billion years ago would be observed as a Gamma ray burst. All calculus on 1/3 page Fundamental theorem of Calculus: DSf(x) = f(x) and SDf(x) = f(x) −f(0). Differentiation rules Dxn = nxn−1 Dea·x = aea·x D cos(a · x) = −a sin(a · x) D sin(a · x) = a cos(a · x) D log(x) = 1/x Integration rules (for x = kh) Sxn = xn+1/(n + 1) Sea·x = (ea·x −1)/a S cos(a · x) = sin(a · x)/a S sin(a · x) = −cos(a · x)/a S 1 x = log(x) Fermat’s extreme value theorem: If Df(x) = 0 and f is continuous, then f has a local maximum or minimum in the open interval (x, x + h). Pictures [x]3 h for h = 0.1 exph(x) for h = 0.1 sinh(x) for h = 0.1 logh(x) for h = 0.1 Here are some worked out examples, similar to what we expect you to do for the homework of lecture 6: The homework should be straightforward, except when finding Sf(x), we want to add a constant such that Sf(0) = 0. In general, you will not need to evaluate functions and can leave terms like sin(5 · x) as they are. If you have seen calculus already, then you could do this exercice by writing d dxf(x) instead of Df(x) and by writing Z x 0 f(x) dx instead of Sf(x). Since we did not introduce the derivative d f/dx nor the integral R x 0 yet, for now, just use the differentiation and integrations rules in the box to the right to solve the problems. 1 Problem: Find the derivative Df(x) of the function f(x) = sin(5 · x) + x7 + 3. Answer: From the differentiation rules, we know Df(x) = 5 cos(5 · x) + 7x6 . 2 Problem: Find the derivative Df(0) of the same function f(x) = sin(5 · x) + 5x7 + 3. Answer: We know Df(x) = 5 cos(5 · x) + 35x6. Plugging in x = 0 gives 5 . 3 Problem: Find the integral Sf(x) of the function f(x) = sin(5 · x) + 5x7 + 3. Answer: From the integration rules, we know Sf(x) = −cos(5 · x)/5 + 5x8/8 + 3x . 4 Problem: Find the integral Sf(1) of the function f(x) = x2 + 1. Answer: From the integration rules, we know Sf(x) = x3/3 + x. Plugging in x = 1 gives 1/3 + 1 if we use the functions in the limit h →0. For positive h, we have to evaluate x(x −h)(x −2h)/3 + x for x = 1 which is (1 −h)(1 −2h)/3 + 1 5 Problem: Find the integral Sf(1) of the function f(x) = exp(4 · x). Answer: From the integration rules, we know Sf(x) = exp(4 · x)/4 −1/4. We have added a constant such that Sf(0) = 0. Plugging in x = 1 gives exp(4)/4 −1/4 . 6 Problem: Assume h = 1/1000. Determine the value of 1 1000[f( 0 1000) + f( 1 1000) + ... + f( 999 1000)] for the function f(x) = −sin(7x) + exp(3x). Answer: The problem asks for Sf(1). We first compute Sf(x) taking care that Sf(0) = 0. Sf(x) = cos(7x)/7 + exp(3x)/3 −(1/7 + 1/3) . Now plug in x = 1 to get cos(7)/7 + exp(3)/3 −(1/7 + 1/3) . Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 6: Worksheet The exponential function We illuminate the fundamental theorem for the exponential function exp(x) = (1+h)x/h. While the discussion could be done for any h > 0 we look at the special case where h = 1 in which case exp(x) = 2x maps positive integers to positive integers. You have verified in a homework that D exp(x) = exp(x) . From the fundamental theorem, we get SD exp(x) = S exp(x) = exp(x) −exp(0) for integers x. That is S exp(x) = exp(x) −1 . In other words, for the exponential function, we know both the deriva-tive and the integral. 1 1 The formula S exp(x) = exp(x) −1 tells for x = 5 that 1 + 2 + 4 + 8 + 16 = 32 −1. Verify it for x = 7. 2 Because S exp(x) = exp(x) −1 we can interpret exp(x) −1 as an area of a union of rectangles. In the picture below, shade an area exp(3) −1. 3 In the right of the two pictures, there is a line vertical segment which has length exp(3). Which one? 1Later in this course, we will look at these two formulas in the limit h →0, where d dx exp(x) = exp(x), Z x 0 exp(t) dt = exp(x) −1 . 4 We know D exp(x) = exp(x). Why is also the following for-mula true? D(exp(x) −1) = exp(x) 5 The just verified formula can be interpreted as a difference between areas and so an area. Which one for x = 4? 0 1 2 3 4 1 2 4 8 16 0 1 2 3 4 1 2 4 8 16 Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 7: Rate of change Last week, we defined Df(x) = f(x + h) −f(x) h . It is the rate of change of the function with step size h. When changing x to x + h and get a response change f(x) to f(x + h). In this lecture, we take the limit h →0 and derive the important formulas d dxxn = nxn−1, d dx exp(x) = exp(x), d dx sin(x) = cos(x), d dx cos(x) = −sin(x) which we have seen already in a discrete setting. 1 You walk up a snow hill of height f(x) = 30−x2 meters. You walk with a step size of h = 0.5 meters. You are at position x = 3. How much do you climb or descend when making an other step? We have f(3) = 21 and f(3.5) = 17.75. We have walked down 3.25 meters. How steep was the snow hill at this point? We have to divide the height difference by the walking distance: −3.25/0.5 = −7.5. Today, we take the limit h →0: If the limit d dxf(x) = limh→0 f(x+h)−f(x) h exist, we say f is differentiable at the point x. The value is called the derivative or instantaneous rate of change of the function f at x. We denote the limit also with f ′(x). 2 In the previous problem, f(x) = 30 −x2 we have f(x + h) −f(x) = [30 −(x + h)2] −[30 −x2] = −2xh −h2 Dividing this by h gives −2x −h. The limit h →0 gives −2x. We have just seen that for f(x) = x2, we get f ′(x) = −2x. For x = 3, this is −6. The actual slope of the snow hill is a bit smaller than the estimate done by walking. The reason is that the hill gets steeper. The derivative f ′(x) has a ge-ometric meaning. It is the slope of the tangent at x. This is an important geometric in-terpretation. It is useful to think about x as ”time” and the derivative as the rate of change of the quantity f(x) in time. -fHxL fHx+hL h x x+h For f(x) = xn, we have f ′(x) = nxn−1. Proof: f(x + h) −f(x) = (x + h)n = (xn + nxn−1h + a2h2 + ... + hn) −xn = nxn−1h + a2h2 + ... + hn). If we divide by h, we get nxn−1 + h(a2 + . . . + hn−2) for which the limit h →0 exists: it is nxn−1. This is an important result because most functions can be approximated very well with polynomials. 3 For f(x) = sin(x) we have f ′(0) = 1 because the dif-ferential quotient is [f(0 + h) −f(0)]/h = sin(h)/h = sinc(h). We have already seen that the limit is 1 before. Lets look at it again geometrically. For all 0 < x < π/2 we have sin(x) ≤x ≤tan(x) . [ dividing by 2 squeezes the area of the sector by the area of triangles.] Because tan(x)/ sin(x) = 1/ cos(x) →1 for x →0, the value of sinc(x) = sin(x)/x must go to 1 as x →0. Renaming the variable x with the variable h, we see the fundamental theorem of trigonometry limh→0 sin(h) h = 1 cosHxL sinHxL tanHxL x x 1 4 For f(x) = cos(x) we have f ′(x) = 0. To see this, look at f(0 + h) −f(0) = cos(h) −1. Geometrically, we can use Pythagoras sin2(h)+(1−cos(h))2 ≤h2 to see that 2−2 cos(h) ≤h2 or 1 −cos(h) ≤h2/2 so that (1 −cos(h))/h ≤h/2 and this goes to 0 for h →0. We have just nailed down an other important identity limh→0 (1−cos(h) h = 0 . The interpretation is that the tangent is horizontal for the cos function at x = 0. We will call this a critical point later on. 5 From the previous two examples, we get cos(x+h)−cos(x) = cos(x) cos(h)−sin(x) sin(h)−cos(x) = cos(x)(cos(h)−1)−sin(x) sin(h) because (cos(h) −1)/h →0 and sin(h)/h →1, we see that [cos(x + h) −cos(x)]/h → −sin(x). For f(x) = cos(ax) we have f ′(x) = −a sin(ax). 6 Similarly, sin(x+h)−sin(x) = cos(x) sin(h)+sin(x) cos(h)−sin(x) = sin(x)(cos(h)−1)+cos(x) sin(h) because (cos(h)−1)/h →0 and sin(h)/h →1, we see that [sin(x+h)−sin(x)]/h →cos(x). for f(x) = sin(ax), we have f ′(x) = a cos(ax). e = lim n→∞(1 + 1 n)n Like π, the Euler number e is irrational. Here are the first digits: 2.7182818284590452354. If you want to find an approximation, just pick a large n, like n = 100 and compute (1 + 1/n)n. For n = 100 for example, we see 101100/100100. We only need 101100 and then put a comma after the first digit to get an approximation. Interested why the limit exists: verify hat the fractions An = (1 + 1/n)n increase and Bn = (1 + 1/n)(n+1) decrease. Since Bn/An = (1 + 1/n) which goes to 1 for n →∞, the limit exists. The same argument shows that (1+1/n)xn = exp1/n(x) increases and exp1/n(x)(1+1/n) decreases. The limiting function exp(x) = ex is called the exponential function. Remember that if we write h = 1/n, then (1 + 1/n)nx = exph(x) considered earlier in the course. We can sandwich the exponential function between exph(x) and (1 + h) exph(x): exph(x) ≤exp(x) ≤exph(x)(1 + h), x ≥0 . For x < 0, the inequalities are reversed. 7 Lets compute the derivative of f(x) = ex at x = 0. Answer. We have ((1 + 1 n)n −1)n ≤(eh −1)/h ≤((1 + 1 n)n+1 −1)n Use the binominal formula to see that both the left and right hand side go to 1 if n →∞. Therefore f ′(0) = 1. The exponential function has a graph which has slope 1 at x = 0. 8 Now, we can get the general case. It follows from ex+h −ex = ex(eh −1) that the derivative of exp(x) is exp(x). For f(x) = exp(ax), we have f ′(x) = a exp(ax). It follows from the properties of taking limits that (f(x) + g(x))′ = f ′(x) + g′(x). We also have (af(x))′ = af ′(x). From this, we can now compute many derivatives 9 Find the slope of the tangent of f(x) = sin(3x) + 5 cos(10x) + e5x at the point x = 0. Solution: f ′(x) = 3 cos(3x) −50 sin(10x) + 5e5x. Now evaluate it at x = 0 which is 3 + 0 + 5 = 8. Finally, lets mention an example of a function which is not everywhere differentiable. 10 The function f(x) = |x| has the properties that f ′(x) = 1 for x > 0 and f ′(x) = −1 for x =< 1. The derivative does not exist at x = 0 evenso the function is continuous there. You see that the slope of the graph jumps discontinuously at the point x = 0. For a function which is discontinuous at some point, we don’t even attempt to differentiate it there. For example, we would not even try to differentiate sin(4/x) at x = 0 nor f(x) = 1/x3 at x = 0 nor sin(x)/|x| at x = 0. Remember these bad guys? To the end, you might have noticed that in the boxes, more general results have appeared, where x is replaced by ax. We will look at this again but in general, the relation f ′(ax) = af(ax) holds (”if you drive twice as fast, you climb twice as fast”). Homework 1 For which of the following functions does the derivative exist for all points? a) | sin(x)| d) | cos(x)| b) | exp(x)| e) sin(1/x) c) exp(x) + sin(15x) f) | exp(x)| + |1 + sin(15x)| 2 a) A circle of radius x has the area f(r) = πr2. Find d drf(r). Can you visualize why this is the same than the circumference of the circle. b) The sphere of radius r has the volume f(r) = 4πr3/3. Find d drf(r)) and compare it with the surface area of the sphere. c) A hypersphere of radius r has the hyper volume f(r) = π2r4/2. Find d drf(r), the volume of the boundary sphere. 3 Find the derivatives of the following functions at the point x = 2. a) f(x) = exp(x) + sin(x) + x + x2 + x3 + x4 + x5. b) f(x) = (x5 −1)/(x −1) + cos(2x). First heal the function. c) f(x) = 1+4x+6x2+4x3+x4 x2+2x+1 . Dito, first heal! 4 In this problem we compute the derivative of √x for x > 0. To do so, we have to find the limit lim h→0 √ x + h −√x h . Hint: multiply the top and the bottom with ( √ x + h + √x) and simplify. 5 A rocket lifts offfrom Cape Canaveral. The height at time t is h(t) = et −1 + √ t, at least for the first few seconds. Find the rate of change of the height at time t = 1. Use the previous problem to get the derivative of √ t. 0.5 1.0 1.5 2.0 2.5 3.0 5 10 15 20 x + ãx - 1 Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 7: Worksheet Rate of change We compute the derivative of f(x) = 1/x by taking limits. a) Simplify 1 x+h −1 x. b) Now take the limit 1 h[ 1 x+h −1 x] when h →0. c) Is there any point where f ′(x) > 0? Derivatives Differentiation rules d dxxn = nxn−1 eax = aeax d dx cos(ax) = −a sin(ax) d dx sin(ax) = a cos(ax) 1 Find the derivatives of the function f(x) = sin(3x) + x5 2 Find the derivative of f(x) = cos(7x) −8x4. 3 Find the derivative of f(x) = e5x + cos(2x). Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 8: The derivative function In the last lecture, we have introduced the derivative f ′(x) = d dxf(x) as a limit of Df(x) for h →0. We have seen that d dxxn = nxn−1 holds for integer n. We also know already that sin′ = cos, cos′ = −sin and exp′ = exp. We can already differentiate a lot of functions and evaluate the derivative f ′(x) at some point x. This is the slope of the curve at x. 1 Find the derivative f ′(x) of f(x) = sin(πx) + cos(πx) −√x + 1/x + x4 and evaluate it at x = 1. Solution: f ′(x) = π cos(πx) −π sin(πx) −1/(2√x −1/x2 + 4x3. Plugging in x = 1 gives −π −1/2 −1 + 4. Taking the derivative at every point defines a new function, the derivative function. For example, for f(x) = sin(x), we get f ′(x) = cos(x). In this lecture, we want to understand the new function and its relation with f. What does it mean if f ′(x) > 0. What does it mean that f ′(x) < 0. Do the roots of f tell something about f ′ or do the roots of f ′ tell something about f? Here is an example of a function and its derivative. Can you see the relation? To understand the relation, it is good to distinguish intervals, where f(x) is increasing or decreasing. This are the intervals where f ′(x) is positive or negative. A function is called monotonically increasing on an interval I = (a, b) if f ′(x) > 0 for all x ∈(a, b). It is monotonically decreasing if f ′(x) < 0 for all x ∈(a, b). Lets look at the previous example again. Here is an interesting inverse problem called bottle calibration problem. We fill a circular bottle or glass with constant amount of fluid. Plot the height of the fluid in the bottle at time t. Assume the radius of the bottle is f(z) at height z. Can you find a formula for the height g(t) of the water? This is not so easy. But we can find the rate of change g′(t). Assume for example that f is constant, then the rate of change is constant and the height of the water increases linearly like g(t) = t. If the bottle gets wider, then the height of the water increases slower. There is definitely a relation between the rate of change of g and f. Before we look at this more closely, lets try to match the following cases of bottles with the graphs of the functions g qualitatively. 2 In each of the bottles, we call g the height of the water level at time t, when filling the bottle with a constant stream of water. Can you match each bottle with the right height function? a) b) d) c) 0.2 0.4 0.6 0.8 1.0 2 4 6 8 10 1) 0.2 0.4 0.6 0.8 1.0 0.1 0.2 0.3 2) 0.2 0.4 0.6 0.8 1.0 -2 2 4 6 8 3) 0.2 0.4 0.6 0.8 1.0 0.1 0.2 0.3 0.4 4) The key is to look at g′(t), the rate of change of the height function. Because [g(t+h)−g(t)] times the area πf 2 is a constant times the time difference h = dt, we have g′ = 1 πf 2 . This formula relates the derivative function of g with the thickness f(t) of the bottle at height g. It tells that if f is large, then g′ is small and if f is small, then g′ is large. Finding g from f is possible but we are not doing this now. 3 Can you find a function f which is bounded |f(x)| ≤1 and such that f ′(x) is unbounded? s Given the function f(x), we can define g(x) = f ′(x) and then take the derivative g′ of g. This second derivative f ′′(x) is called the acceleration. It measures the rate of change of the tangent slope. For f(x) = x4, for example we have f ′′(x) = 12x2. If f ′′(x) > 0 on some interval the function is called concave up, if f ′′(x) < 0, it is concave down. 4 Find a function f which has the property that its acceleration is constant equal to 10. 5 Can you find a function f which is bounded |f(x)| ≤1 and such that f ′′(x) is positive everywhere? Homework 1 For the following functions, determine on which intervals the function is monotonically in-creasing or decreasing. a) f(x) = x3 −x. b) f(x) = sin(πx). c) f(x) = e2x −2ex 2 Match the following functions with their derivatives. Give short explanations for each match. a) b) c) d) 1) 2) 3) 4) 3 Match also the following functions with their derivatives. Give short explanations docu-menting your reasoning in each case. a) b) c) d) 1) 2) 3) 4) 4 Draw for the following functions the graph of the function f(x) as well as the graph of its derivative f ′(x). You do not have to compute the derivative analytically as a formula here since we do not have all tools yet to compute the derivatives. The derivative function you draw needs to have the right qualitative shape however. a) The Gaussian bell curve or the ”To whom the bell tolls” function f(x) = e−x2 b) The witch of Maria Agnesi. f(x) = 1 1 + x2 c) The three gorges function f(x) = 1 x + 1 x −1 + 1 x + 1 . 5 Below you the graphs of three derivative functions f ′(x). In each case you are told that f(0) = 1. Your task is to draw the function f(x) in each of the cases a),b),c). Your picture does not have to be up to scale, but your drawing should display the right features. a) b) c) Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 8: Worksheet Matching functions with their derivative 0) o) O) In this worksheet we want to match the graphs of functions with their deriva-tives and second derivatives. This is tougher than you might think. Here is an example: The first graph shows the function, which is here the quadratic function. The slope on the right hand side is pos-itive and increasing, on the left hand side the function is negative and de-creasing. The middle graph shows the derivative function which is linear. The final graph shows the derivative function of the derivative function. It is constant in this case. 1 Match the following functions with their derivatives and then with the derivatives of the derivatives. a) b) c) d) 1) 2) 3) 4) A) B) C) D) Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 9: The product rule In this lecture, we look at the derivative of a product of func-tions. The product rule is also called Leibniz rule named after Gottfried Leibniz, who found it in 1684. It is a very important rule because it allows us differentiate many more functions. If we wanted to compute the derivative of f(x) = x sin(x) for example, we would have to get under the hood of the function and compute the limit lim(f(x + h) −f(x))/h. We are too lazy for that. Lets start with the identity f(x + h)g(x + h) −f(x)g(x) = [f(x + h) −f(x)] · g(x + h) + f(x) · [g(x + h) −g(x)] which can be written as D(fg) = Dfg+ + fDg with g+(x) = g(x + h). This quantum Leibniz rule can also be seen geometrically: the rectangle of area (f + d f)(g + dg) is the union of rectangles with area f · g, f · dg and d f · g+. Divide this relation by h to see [f(x + h) −f(x)] h · g(x + h) → f ′(x) · g(x) f(x) · [g(x + h) −g(x)] h → f(x) · g′(x) . We get the extraordinarily important product rule: f g f dg + dfg d dx(f(x)g(x)) = f ′(x)g(x) + f(x)g′(x) . 1 Find the derivative function f ′(x) for f(x) = x3 sin(x). Solution: We know how to differ-entiate x3 and sin(x) so that f ′(x) = 3x2 sin(x) + x3 cos(x). 2 While we know d dxx5 = 5x4 lets compute this with the Leibniz rule and write x5 = x3 · x2. We have d dxx3 = 3x2, d dxx2 = 2x . The Leibniz rule gives us d/dx5 = 3x4 + 2x4 = 5x4. 3 Water powered JetLev systems have now gone into produc-tion. The water is sucked up from the water surface from a four-inch diameter polyester hose. Consider a system, where the water is carried with you. By Newtons law the force F satisfies F = p′, where p = mv is the momentum, the product of your mass and velocity. Written out, this is F(t) = d dt(m(t)v(t)) . How big is the acceleration v′? The product rule tells us F = m′v + mv′ which gives v′ = (F −m′v)/m. Since we throw out water, m′(t) is negative and m(t) decreases, we accelerate if the force F is kept constant. The Leibniz rule is also called product rule. It suggests a quotient rule. One can avoid the quotient rule by writing it as a product f(x)/g(x) = f(x) · 1/g(x) and by using the reciprocal rule: If g(x) ̸= 0, then d dx 1 g(x) = −g′(x) g(x)2 . To verify it, stare at the identity 1 g(x + h) − 1 g(x) = g(x) −g(x + h) g(x)g(x + h) . Dividing it by h gives D(1/g(x)) = −Dg(x)/(g(x)g+(x)). Taking the limit h →0 leads to the identity. An other way to derive this is to write h = 1/g and differentiate 1 = gh on both sides. The product rule gives 0 = g′h + gh′ so that h′ = −hg′/g = −g′/g2. 4 Find the derivative of f(x) = 1/x4. Solution: f ′(x) = −4x3/x8 = −4/x5. The same computation shows that d dxxn = nxn−1 holds for all integers n. The formula d dxxn = nxn−1 holds for all integers n. The quotient rule is obtained by applying the product rule to f(x) · (1/g(x)) and using the reciprocal rule: If g(x) ̸= 0, then d dx f(x) g(x) = [f ′(x)g(x) −f(x)g′(x)] g2(x) . 5 Find the derivative of f(x) = tan(x). Solution: because tan(x) = sin(x)/ cos(x) we have tan′(x) = sin2(x) + cos2(x) cos2(x) = 1 cos2(x) . 6 Find the derivative of f(x) = 2−x x2+x4+1. Solution. We apply the quotient rule and get [(−1)x2 + x4 + 1 + (2 −x)(2x + 4x3)]/(x2 + x4 + 1). Here are some more problems with solutions: 7 Find the second derivative of tan(x). Solution. We have already computed tan′(x) = 1/ cos2(x). Differentiate this again with the quotient rule gives −d dx cos2(x) cos4(x) . We still have to find the derivative of cos2(x). The product rule gives cos(x) sin(x) + sin(x) cos(x) = 2 cos(x) sin(x). Our final result is 2 sin(x)/ cos3(x) . 8 A cylinder has volume V = πr2h, where r is the radius and h is the height. Assume the radius grows like r(t) = 1+t and the height shrinks like 1−sin(t). Does the volume grow or decrease at t = 0? Solution: The volume V (t) = π(1 + t)2(1 −sin(t)) is a product of two functions f(t) = π(1 + t)2 and g(t) = (1 −sin(t). We have f(0) = 1, g′(0) = 2, f ′(0) = 2, g(0) = 1. The product rule gives gives V ′(0) = π1 · (−1) + π2 · 1 = π. The volume increases in volume at first. 9 On the Moscow papyrus dating back to 1850 BC, the general formula V = h(a2+ab+b2)/3 for a truncated pyramid with base length a, roof length b and height h appeared. Assume h(t) = 1 + sin(t), a(t) = 1 + t, b(t) = 1 −2t. Does the volume of the truncated pyramid grow or decrease at first? Solution. We could fill in a(t), b(t), h(t) into the formula for V and compute the derivative using the product rule. A bit faster is to write f(t) = a2 + ab + b2 = (1+t)2+(1−3t)2+(1+t)(1−3t) and note f(0) = 3, f ′(0) = −6 then get from h(t) = (1+sin(t)) the data h(0) = 1, h′(0) = 1. So that V ′(0) = (h′(0)f(0)−h(0)f ′(0))/3 = (1·3−1(−6))/3 = −1. The pyramid shrinks in volume at first. 10 We pump up a balloon and let it fly. Assume that the thrust increases like t and the resistance decreases like 1/√1 −t since the balloon gets smaller. The distance traveled is f(t) = t/√1 −t. Find the velocity f ′(t) at time t = 0. Homework 1 Find the derivatives of the following functions: a) f(x) = sin(3x) cos(10x). b) f(x) = sin2(x)/x2. c) f(x) = x4 sin(x) cos(x). d) f(x) = 1/√x. e) f(x) = cot(x) + (1 + x)/(1 + x2). 2 a) Verify that for f(x) = g(x)h(x)k(x) the formula f ′ = g′hk + gh′k + ghk′ holds. b) Verify the following formula for derivative of f(x) = g(x)3: f ′(x) = 3g2(x)g′(x). 3 a) If f(x) = sinc(x) = sin(x)/x, find its derivative g(x) = f ′(x) and then the derivative of g(x). Then evaluate it at x = 0. b) If you evaluate g(x) at x = 0 you obtain g(0) = f ′(0) = 0. Is the result in a) not a contradiction to the fact that for g = 0 the derivative g′ is 0? 4 Find the derivative of sin(x) 1 + cos(x) + x4 1+cos2(x) at x = 0. 5 a) Verify that in general the derivative of g(x) = f(x)2 is 2f(x)f ′(x). b) We have already computed the derivative of f(x) = √x in the last homework by directly computing the limit. Lets do it using the product rule. Use part a) of this problem to compute the derivative of g(x) = f(x) · f(x) Use the obtained identity g′(x) = ... to get a formula for f ′(x) = d dx q g(x). c) Use the same method and the above homework problem 2b) in this homework set to compute the derivative of the cube root function f(x) = x1/3. This last problem 5) is a preparation for the chain rule, we see next Monday. Avoid using the chain rule already here. Remarks: Like quantum calculus also quantum Leibniz rule is old. . The obove picture explaining the discrete rule (without having to consider any error terms) appears in the article John Dawson, ”Wavefronts, BoxDiagrams and the Product Rule: A discovery Approach”, 11 Page 102-106, Two Year College Mathematics Journal, 1980. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 9: Worksheet The product rule We practice the product, reciprocal and quotient rule 1 Find the derivative of the sinc-function sin(x)/x at the point x = 0. 2 What is the slope of the graph of the function f(x) = xe−x2 at x = 0? 3 Find the derivative of √x/x at x = 1. (Look first!) 4 Find the derivative of 1/ex at x = 1. (Look first!) 5 Assume we remember the formula sin(2x) = 2 sin(x) cos(x). Differentiate both sides to get a formula for cos(2x). 6 Find the derivative of x −1/(x2 + 1) at x = 0. Source: XKCD Leibniz 1684 paper. The product and quotient rule is introduced. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 10: The chain rule In this lecture, we look at the derivative of a composition of functions. Also this rule is important. It will allow us to compute derivatives like for f(x) = sin(x3) which is a composition of two functions f(x) = x3 and g(x) = sin(x). We can in this example not use the product rule since we do not have a product of functions. It is a composition of functions. How do we compute the derivative functions which are ”chained” together like this? The answer to this question is given by the chain rule: d dxf(g(x)) = f ′(g(x))g′(x) . The chain rule follows from the identity f(g(x + h)) −f(g(x)) h = [f(g(x) + (g(x + h) −g(x))) −f(g(x))] [g(x + h) −g(x)] · [g(x + h) −g(x)] h . Write H(x) = g(x+h)-g(x) in the first part on the right hand side f(g(x + h)) −f(g(x)) h = [f(g(x) + H) −f(g(x))] H · g(x + h) −g(x) h . As h →0, we also have H →0 and the first part goes to f ′(g(x)) and the second factor has g′(x) as a limit. 1 Find the derivative of f(x) = (4x −1)17. Solution The inner function is 4x −1 which has the derivative 4. We get therefore f ′(x) = 17(4x −1)6 · 4 = 28(4x −1)6. Remark. We could have expanded out the power (4x−1)17 first and avoided the chain rule. Avoiding the chain rule is called the pain rule . 2 Find the derivative of f(x) = sin(π cos(x)) at x = 0. Solution: applying the chain rule gives cos(π cos(x)) · (−π sin(x)). 3 For linear functions f(x) = ax + b, g(x) = cx + d, the chain rule can readily be checked. We have f(g(x)) = a(cx + d) + b = acx + ad + b which has the derivative ac. Indeed this is the definition of f times the derivative of g. You can convince you that the chain rule is true also from this example since if you look closely at a point, then the function is close to linear. One of the cool applications of the chain rule is that we can compute derivatives of inverse functions: 4 Find the derivative of the natural logarithm function log(x) 1 Solution Differentiate the identity exp(log(x)) = x. On the right hand side we have 1. On the left hand side the chain rule gives exp(log(x)) log′(x) = x log′(x) = 1. Therefore log′(x) = 1/x. 1We always write log(x) for the natural log. The ln notation is old fashioned and only used in obscure places like calculus books and calculators from the last millenium. d dx log(x) = 1/x. Denote by arccos(x) the inverse of cos(x) on [0, π] and with arcsin(x) the inverse of sin(x) on [−π/2, π/2]. 5 Find the derivative of arcsin(x). Solution. We write x = sin(arcsin(x)) and differentiate. d dx arcsin(x) = 1 √ 1 −x2 . 6 Find the derivative of arccos(x). Solution. We write x = cos(arccos(x)) and differentiate. d dx arccos(x) = − 1 √ 1 −x2 . 7 f(x) = sin(x2 + 3). Then f ′(x) = cos(x2 + 3)2x. 8 f(x) = sin(sin(sin(x))). Then f ′(x) = cos(sin(sin(x))) cos(sin(x)) cos(x). Why is the chain rule called ”chain rule”. The reason is that we can chain even more functions together. 9 Lets compute the derivative of sin( √ x5 −1) for example. Solution: This is a composition of three functions f(g(h(x))), where h(x) = x5 −1, g(x) = √x and f(x) = sin(x). The chain rule applied to the function sin(x) and √ x5 −1 gives cos( √ x5 −1) d dx √ x5 −1. Apply now the chain rule again for the derivative on the right hand side. 10 Here is the famous falling ladder problem. A stick of length 1 slides down a wall. How fast does it hit the floor if it slides horizontally on the floor with constant speed? The ladder connects the point (0, y) on the wall with (x, 0) on the floor. We want to express y as a function of x. We have y = f(x) = √ 1 −x2. Taking the derivative, assuming x′ = 1 gives f ′(x) = −2x/ √ 1 −x2. 1 x y In reality, the ladder breaks away from the wall. One can calculate the force of the ladder to the wall. The force becomes zero at the break-away angle θ = arcsin((2v2/(3g))2/3), where g is the gravitational acceleration and v = x′ is the velocity. 11 For the brave: find the derivative of f(x) = cos(cos(cos(cos(cos(cos(cos(x))))))). Homework 1 Find the derivatives of the following functions a) f(x) = cos(√x b) f(x) = tan(1/x5) c) f(x) = exp(1/(1 + x)) d) (2 + sin(x))−5 2 Find the derivatives of the following functions at x = 1. (Problems c),d) were cut off on the originally distributed pset and are not required. Do them nevertheless to have more practice). a) f(x) = x4 log(x). b) √ x5 + 1 c) (1 + x2 + x4)100 d) 5x4 2 √ x5+1 3 a) Find the derivative of f(x) = 1/x by differentiating the identity xf(x) = 1. b) Find the derivative of f(x) = arccot(x) by differentiating cot(arccot(x)) = x. 4 a) Find the derivative of f(x) = √x by differentiating the identity f(x)2 = x. b) Find the derivative of f(x) = xm/n by differentiating the identity f(x)n = xm. The function f(x) = [exp(x) + exp(−x)]/2 is called cosh(x). The function f(x) = [exp(x) −exp(−x)]/2 is called sinh(x). They are called hyperbolic cosine and hyperbolic sine. The first is even, the sec-ond is odd. You can see directly using exp′(x) = exp(x) and exp′(−x) = −exp(−x) that sinh′(x) = cosh(x) and cosh′(x) = sinh(x). Furthermore exp = cosh + sinh writes exp as a sum of an even and odd function. 5 a) Find the derivative of the inverse arccosh(x) of cosh(x). b) Find the derivative of the inverse arcsinh(x) of sinh(x). coshHxL x sinhHxL x Apropos chain: if you look at the shape of a chain hanging at two points, then it is in the shape of the hyperbolic cosine. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 10: Worksheet The chain rule On this valentine day, we preview a nice application of the chain rule. We will cover this later in the course. The Valentine equation (x2 + y2 −1)3 −x2y3 = 0 relates x with y, but we can not write the curve as a graph of a function y = g(x). Extracting y or x is difficult since they are in love. The set of points satisfying the equation looks like a heart. Well, romance is known to be complicated! You can check that (1, 1) satisfies the Valentine equation. Near it, the curve looks like the graph of a function g(x). Lets fill that in and look at the function f(x) = (x2 + g(x)2 −1)3 −x2g(x)3 The key is that f(x) is actually zero and if we take the derivative, then we get zero too. Using the chain rule, we can take the derivative f ′(x) = 3(x2+g(x)2−1)(2x+2g(x)g′(x))−2xg(x)3−x23g(x)2g′(x) = 0 Magically, we can solve for g′ g′(x) = − 3(x2 + g(x)2 −1)2x −2xg(x)3 3(x2 + g(x)2 −1)2g(x) −3x2g(x)2 . Filling in x = 1, g(x) = 1, we see this is −4/3. We have computed the slope of g without knowing g. Isn’t that magic? If this was a bit too complicated, don’t worry. We will have an entire lecture on this later in the course. 1 Compute the derivative of f(x) using the chain rule and verify the formula above. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 11: Local extrema Maximizing and minimizing functions is an important task. The reasons are obvious: we want to maximize nice quantities and minimize unpleasant ones. Extremizing quantities is also the most important principle nature follows. Important laws in physics like Newtons law, equations describing light, or matter can be based on the principle of extremization. The intuition is that at maxima or minima the tangent to the graph is horizontal. This leads to a zero derivative and the notion of critical points: A point x0 is a critical point of a differentiable function f if f ′(x0) = 0. In some textbooks, critical points include points where f ′ is not defined. In this course we do not include these points in the list of critical points. They are points outside the domain of definition of f ′ and will be treated separately. 1 Find the critical points of the function f(x) = x3 + 3x2 −24x. Solution: we compute the derivative as f ′(x) = 3x2 + 6x −24. The roots of f ′ are 2, −4. A point is called a local maximum of f, if there exists a neighborhood U = (p −a, p + a) of p, such that f(p) ≥f(x) for all x ∈U. Similarly, we define a local minimum. Local maxima and minima together are called local extrema. 2 The point x = 0 is a local maximum for f(x) = cos(x). The reason is that f(0) = 1 and f(x) < 1 nearby. 3 The point x = 1 is a local minimum for f(x) = (x −1)2. The function is zero at x = 1 and positive everywhere else. Fermat: If f is differentiable and has a local extremum at x, then f ′(x) = 0. Why? Assume the derivative f ′(x) = c is not zero. We can assume c > 0 other-wise replace f with −f. By the definition of limits, for some large enough h, we have f(x + h) −f(x)/h ≥c/2. But this means f(x + h) ≥f(x) + hc/2 and x can not be a local maximum. Since also (f(x) −f(x −h))/h ≥c/2 for small enough h, we also have f(x −h) ≤f(x) −hc/2 and x can not be a local minimum. 1 2 4 The derivative of f(x) = 72x−30x2−8x3+3x4 is f ′(x) = 72−60x−24x2 +12x3 By plugging in integers (calculus teachers like integer roots because students like integer roots!) we can guess the roots x = 1, x = 3, x = −2 and see f ′(x) = 12(x −1)(x + 2)(x −3). The critical points are 1, 3, −2. 5 We have already seen that f ′(x) = 0 does not assure that x is a local extremum. The function f(x) = x3 is a counter example. It satisfies f ′(0) = 0 but 0 is not a local extremum. It is an example of an inflection point, a point where f ′′ changes sign. 6 Lets look at one nasty example. The function f(x) = x sin(1/x) is continuous at 0 but there are infinitely many critical points near 0. If f ′′(x) > 0, then the graph of the function is concave up. If f ′′(x) < 0 then the graph of the function is concave down. Second derivative test. If x is a critical point of f and f ′′(x) > 0, then f is a local minimum. If f ′′(x) < 0, then f is a local maximum. If f ′′(x0) > 0 then f ′(x) is negative for x < x0 and positive for f ′(x) > x0. This means that the function decreases left from the critical point and increases right from the critical point. Similarly, if f ′′(x0) < 0 then f ′(x) is positive for x < x0 and f ′(x) is positive for x > x0. This means that the function increases left from the critical point and increases right from the critical point. 7 The function f(x) = x2 has one critical point at x = 0. Its second derivative is 2 there. 8 Find the local maxima and minima of the function f(x) = x3 −3x using the second derivative test. Solution: f ′(x) = 3x2 −3 has the roots 1, −1. The second derivative f ′′(x) = 6x is negative at x = −1 and positive at x = 1. The point x = −1 is therefore a local maximum and the point x = 1 is a local minimum. 9 Find the local maxima and minima of the function f(x) = cos(πx) using the second derivative test. 10 For the function f(x) = x5 −x3, the second derivative test is inconclusive at x = 0. Can you nevertheless see the critical points? 3 11 Also for the function f(x) = x4, the second derivative test is inconclusive at x = 0. The second derivative is zero. Can you nevertheless see whether the critical point 0 is local maximum of local minimum? Finally, lets look at an example, where we can practice a bit the chain rule. 12 Find the critical points of f(x) = 4 arctan(x) + x2. Solution. The derivative is f ′(x) = 4 1 + x2 + 2x = 2x + 2x3 + 4 1 + x2 . We see that x = −1 is a critical point. There are no other roots of 2x + 2x3 + 4 = 0. How did we get the derivative of arctan again? Differentiate tan(arctan(x)) = x and write u = arctan(x) : 1 cos2( u ) arctan′(x) = 1 . Use the identity 1 + tan2( u ) = 1/ cos2( u ) to write this as (1 + tan2( u )) arctan′(x) = 1 . But tan( u ) = tan( arctan(x) )= x so that tan2(u) = x2. And we have (1 + x2) arctan′(x) = 1 . Now solve for arctan′(x): arctan′(x) = 1 1 + x2 . 4 Homework 1 Find all critical points for the following functions. If there are infinitely many, indicate their structure. For f(x) = cos(x) for example, the critical points can be written as π/2 + kπ, where k is an integer. a) f(x) = x4 −3x2. b) f(x) = 3 + sin(πx) c) f(x) = exp(−x2)x2. d) f(x) = cos(sin(x)) 2 For the following functions, find all the maxima and minima using the second deriva-tive test: a) f(x) = x log(x), where x > 0. b) f(x) = 1/(1 + x2) c) f(x) = x2 −2x + 1. d) f(x) = 2x tan(x), where −π/2 < x < π/2 3 Verify that a cubic equation f(x) = x3 + ax2 + bx + c always has an inflection point, a point where f ′′(x) changes sign. Hint. Remember the wobbling table! 4 Depending on c, the function f(x) = x4 −cx2 has either one or three critical points. Find these points for a general c and use the second derivative test to see whether they are maxima or minima. The answer will depend on c. Where does the answer change? 5 This creative problem is motivated from an interesting observation of Kent done last week in class. You can write down explicit formulas (of course you can experiment with graphing software) or just draw the graph. If you think no solution exists, indicate so. a) Find a function which has exactly 2 local maximum and 1 local minimum. b) Find a function which has exactly 2 local maxima and no local minimum. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 11: Worksheet Critical points and extrema Which rectangle of fixed area xy = 1 has minimal circumference 2x+ 2y? We have to extremize the function f(x) = 2x + 2 x . 1 Differentiate the function f. For which x is it continuous? 2 Find the critical points of f, the places where f ′(x) = 0. 3 Sketch the graph of f on the interval (0, 4]. x y=1 x y 0 4 x fHxL A related but much more difficult problem is to find the shape with fixed area 1 which has minimal circumference. A more advanced flavor of calculus allows to solve this: the calculus of variations. 1 So, which is the winner? You might have guessed. The circle. The rectangle example illustrates already that symmetry is often favored in extremization problems. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 12: Global extrema In this lecture we look at super maxima. Local maxima are great, global maxima are the greatest. These extrema can occur at critical points of f or at the boundary of the domain, where f is defined. A point p is called a global maximum of f if f(p) ≥f(x) for all x. A point p is called a global minimum of f if f(p) ≤f(x) for all x. How do we find global maxima? We just make a list of all local extrema and boundary points, then pick the largest. Global extrema do not need to exist on the real line. The function f(x) = x2 has a global minimum at x = 0 but no global maximum. We can however look at global maxima on finite intervals. 1 Find the global maximum of f(x) = x2 on the interval [−1, 4]. Solution. We look for local extrema at critical points and at the boundary. Then we compare all these extrema to find the maximum or minimum. The critical points are x = 0. The boundary points are −1, 4. Comparing the values f(−1) = 1, f(0) = 0 and f(4) = 16 shows that f has a global maximum at 4 and a global minimum at 0. Extreme value theorem A continuous function f on a finite interval [a, b] attains a global maximum and a global minimum. Here is the argument: Because the function is continuous, the image f([a, b]) is a closed interval [c, d]. 1 There is a point such that f(x) = c, which is a global minimum and a point where f(x) = d which is a global maximum. Note that the global maximum or minimum can also also on the boundary or points where the derivative dos not exist. 2 Find the global maximum and minimum of the function f(x) = |x|. The function has no absolute maximum as it goes to infinity for x →∞. The function has no critical point on the domain of definition R \ {0} of the function f ′. To see the minimum, we have also to look at the point x = 0. 3 A soda can is a cylinder of volume πr2h. The surface area 2πrh + 2πr2 measures the amount of material used to manufacture the can. Assume the surface area is 2π, we can solve the equation for h = (1 −r2)/r = 1/r −r Solution: The volume is f(r) = π(r−r3). Find the can with maximal volume: f ′(r) = π −3r2π = 0 showing r = 1/ √ 3. This leads to h = 2/ √ 3. 1This statement needs more justification but is intuitive enough that we can accept it. 1 2 4 Take a US Letter size paper of 8×11 inches. 2 If we cut out 4 squares of equal size at the corners, we can fold up the paper to a tray with width (8−2x) length (11−2x) and height x. Find the x ∈[0, 4] for which the volume f(x) = (8 −2x)(11 −2x)x = 4x3 −38x2 + 88x is maximal. The solutions to f ′(x) = 12x2 −76x + 88 = 0 are x = i(19 ± √ 97)/2 which is about 1.5 or 5. The second one is larger than 4. We see that What is the minimal volume? This example illustrates that we might have to look at the boundary of the interval for extrema. Assume we have a function f which is differentiable except at some points a1, . . . , an. We include the end points of the domain of definition in this list. The task is to find the global maximum. How do we proceed? • 1. Evaluate the function at the point a1, .., an. • 2. Find the local maxima by looking at critical points b1, ., ..bn. • 3. Find the maximum of f(a1), f(a2), ..., f(an), f(b1), ...f(bn). 5 Find the global maxima and minima of the function f(x) = |x| −2x2 + x3 on the interval [−1, 2]. 1) The function is differentiable except at x = 0. On x > 0 the function is f(x) = x −2x2 + x3. It has derivative 1 −4x + 3x2 which has the root 1/3, 1. On x < 0 the function is f(x) = −x −2x2 + x3, which has a critical point at (2 − √ 7)/3 = −0.215.. There is an other critical point but that one is above x < 0. So we have the three critical points 1/3, 1, (2 − √ 7)/3. 2) The function is not differentiable at x = 0 and has the boundary points 1, 2. 3) If we evaluate f at the critical points, we get the val-ues (0, 0.148, 0, 0.1125, −2, 2). The global maximum is at x = 2. 2The correct size is 17/2 × 11 inches we avoid fractions. 3 Homework 1 Find the global maxima and minima of the function f(x) = (x −2)2 on the interval [0, 3]. 2 Find the global maximum and minimum of the function f(x) = 2x3 −3x2 −36x on the interval [−5, 5] 3 A candy manufacturer builds spherical candies. Its effectivness is A(r) −V (r), where A(r) is the surface area and V (r) the volume of a candy of radius r. Find the radius, where f(r) = A(r) −V (r) has a global maximum for r ≥0. 4 Lets look at the falling ladder again. But now x denotes the angle, the ladder makes with the floor. Find the angle, where the distance f(x) of the ladder to the wall-floor corner is maximal. P.S. You can assume the ladder has length 1 but it will not matter how long the ladder is. x fHxL 4 5 a) The function S(p) = −p log(p) is called the entropy function. 3 Find the probability 0 < p ≤1 which maximizes entropy. One of the most important principle in all science is that nature tries to maximize entropy. In some sense we compute here the number of maximal entropy. b) We can write 1/xx = e−x log(x). Find the value x, where x−x has a local maximum which is the point where xx has a local minimum. . 4 3If W = 1/p is the ”Wahrscheinlichkeit”, the number of microstates, then S(p) = p log(W) is the expectation of W, also written just as log(W). This relation between probability and entropy is inscribed on Bolzmann’s thombstone S = k log(W), where k is an additional constant which depends on units. The Bolzman entropy formula has far reaching consequences to very concrete problems in chemistry. 4The identity ab = eb log(a) one of the three properties to remember for exponentials. The other two are abac = ab+c and (ab)c = abc. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 12: Worksheet Extrema with boundaries The following famous problem is usually asked with the Statue of liberty. At Harvard, we of course want to use the John Harvard Statue. It is a common situation. You want to look at a statue. If you are too close below it, the viewing angle becomes small. If you are far away, the viewing angle decreases again. There is an optimal distance where the viewing angle is maximal. At which distance x do you see most of the John Harvard Statue? Assume the part you want to see 4 to 9 feet higher than your eyes. 1 Verify that the angle you see from the statue is f(x) = arctan(9 x) −arctan(4 x) . 2 Differentiate f(x) to find the minimum. 3 Are there any boundary points or points where f is not differentiable? 4 Find the global maximum of f. 5 Is there a global minimum of f? Here is a graph of part of the function f. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 13: Hopitals rule The rule In this lecture, we look at a powerful rule to compute limits. This Hopital’s rule works miracles and solves all our remaining worries about limits: Hopital’s rule. If f, g are differentiable and f(p) = g(p) = 0 and g′(p) ̸= 0, then lim x→p f(x) g(x) = f ′(p) g′(p) . Lets see how it works: 1 Lets prove the fundamental theorem of trigonometry again: lim x→0 sin(x) x = lim x→0 cos(x) 1 = 1 . Why did we work so hard for this? Note that we used the fundamental theorem to derive the derivatives for cos and sin at all points. In order to apply l’Hopital, we had to know the derivative. Our work to establish the limit was not in vain. The proof of the rule is almost comic in its simplicity if we compare it with how fantastically useful it is: Since f(p) = g(p) = 0 we have Df(p) = f(p + h)/h and Dg(p) = g(p + h)/h so that for every h > 0 with g(p + h) ̸= 0 the quantum l’Hopital rule holds: f(p + h) g(p + h) = Df(p) Dg(p) . Now take the limit h →0. The left side is what we want to know, the right side is a quotient of two limits which exist since g′(p) ̸= 0. 1 Sometimes, we have to administer a medicine twice. To use this, l’Hopital can be improved in that the condition g′(0) = 0 can be replaced by the requirement that the limit limx→p f ′(x)/g′(x) exists. Instead of having a rule which replaces a limit with an other limit (we cure a disease with a new one!) we formulate it in the way how it is actually used. The second derivative case could easily be generalized for higher derivatives. There is no need to memorize this. Just remember that you can check in several times to a hospital. If f(p) = g(p) = f ′(p) = g′(p) = 0 then limx→p f(x) g(x) = f′′(p) g′′(p) if g′′(p) ̸= 0. 1Some books refer to the intermediate value theorem here. This is not necessary. 1 2 2 Find the limit limx→0(1 −cos(x))/x2. Remember that this limit had also been pivotal to compute the derivatives of trigonometric functions. Solution: differentiation gives lim x→0 −sin(x)/2x . This limit can be obtained with l’Hopital again. lim x→0 −sin(x)/(2x) = lim x→0 −cos(x)/2 = −1/2 . 3 Find the limit f(x) = (exp(x2) −1)/ sin(x2) for x →0. 4 What do you get if you apply l’Hopital to the limit [f(x + h) −f(x)]/h as h →0? 5 Find limx→∞x sin(1/x). Solution. Write y = 1/x then sin(y)/y. Now we have a limit, where the denominator and nominator both go to zero. The case when both sides converge to infinity can be reduced the other case by look-ing at A = f/g = (1/g(x))/(1/f(x)) which has the limit g′(x)/g2(x)/f ′(x)/f 2(x) = g′(x)/f ′(x)((1/g)/(1/f))2 = g′/f ′(f 2/g2) = (g′/f ′)A2, so that A = f ′(p)/g′(p). We see: If limx→p f(x) = limx→p g(x) = ∞for x →p and g′(p) ̸= 0, then lim x→p f(x) g(x) = f ′(p) g′(p) . 2 What is the limit limx→0 xx? This answers the intriguing question: what is What is 00? Solution: Because xx = ex log(x), it is enough to understand the limit x log(x) for x →0. lim x→0 log(x) 1/x . Now the limit can be seen as the limit (1/x)/(−1/x2) = −x which goes to 0. Therefore limx→0 xx = 1. (We assume x > 0 to have real values xx) 3 Find the limit limx→2 x2−4x+4 sin2(x−2). Solution: this is a case where f(2) = f ′(2) = g(2) = g′(2) = 0 but g′′(0) = 2. The limit is f ′′(2)/g′′(2) = 2/2 = 1. Hopital’s rule always works in calculus situations, where functions are differentiable. The rule can fail if differentiability of f or g fails. Here is an other ”rare” example: 4 Deja Vue: Find √ x2+1 x for x →∞. L’Hopital gives x/ √ x2 + 1 which in terms gives again √ x2+1 x . Apply l’Hopital again to get the original function. We got an infinite loop. If the limit is A, then the procedure tells that it is equal to 1/A. The limit must therefore be 1. This case can be covered easily without l’Hopital: divide both sides by x to get p 1 + 1/x2. Now, we can see the limit 1. 5 Scarecrow: given f(x) = x sin(1/x4)e−1/x2 and g(x) = e−1/x2. What is the limit f(x)/g(x) as x →0. Solu-tion. Since the functions f and g are not differentiable at x = 0 l’Hopital is not appropriate. The example ap-pears in textbooks because the limit still exists. Look at f/g = x sin(1/x4) which satisfies |f(x)/g(x)| ≤|x| and converges to 0 for x →0. 2 2It appears in 3 6 Given a differentiable function satisfying g(0) = 0. Verify that the limit limx→0 f(g(x))/g(x) is f ′(0). Solution: You check in the homework that the result is f ′(g(0)). History The rule appeared in the ”first calculus book” the world has known. The book with name ”Analyse des Infiniment Petits pour l”intelligence des Lignes Courbes” appeared in 1696 and was written by Guillaume de l’Hopital, a text if typeset in a modern font would probably fit onto 50-100 pages.3 It is now clear that the mathematical con-tent of Hopital’s book is mostly due to Johannes Bernoulli who became a math-ematical ”mercenary” for l’Hopital: Clifford Truesdell write in his article ”The New Bernoulli Edition”, 4 about this ”most extraordinary agreement in the history of sci-ence”: l’Hopital wrote: ”I will be happy to give you a retainer of 300 pounds, beginning with the first of January of this year ... I promise shortly to increase this retainer, which I know is very modest, as soon as my affairs are somewhat straightened out ... I am not so unreasonable as to demand in return all of your time, but I will ask you to give me at intervals some hours of your time to work on what I request and also to communicate to me your discoveries, at the same time asking you not to disclose any of them to others. I ask you even not to send here to Mr. Varignon or to others any copies of the writings you have left with me; if they are published, I will not be at all pleased. Answer me regarding all this ...” Bernoulli’s response is lost, but a letter from l’Hopital indicates that it was quickly accepted. From this point on, Bernoulli was a ”giant enchained” (Truesdell). Clifford Truesdell also mentions that the book of l’Hopital has remained the standard for Calculus for a century. 3Stewart’s book with 1200 pages probably contains about 4 million characters, about 12 times more than l’Hopital’s book. It also contains more material of course. The OCRed text of l’Hopital’s book of 200 pages has 300’000 characters. 4Isis, Vol. 49, No. 1, 1958, pages 54-62 4 Homework 1 For the following functions, find the limits as x →0: a) (x2 −x)/ sin(x) b) (exp(x) −1)/(exp(3x) −1) c) sin3(3x)/ sin3(5x) d) x + log(x)x + sin(x2) sin2(x) e) sin(sin(sin(exp(sin(x)))))/ sin(sin(exp(sin(x)))). 2 For the following functions, find the limits as x →1: a) (x2 −x −1)/(cos(x −1) −1) b) (exp(x) −e)/(exp(3x) −e3) For the following functions, find the limits as x →∞: c) (x2 −x −1)/ √ x4 + 1 d) (x −4)/(4x + sin(x) + 8) 3 Here is an FUD attempt on l’Hopitals rule: Define f(x) = x + cos(x) sin(x) and g(x) = exp(sin(x))(x + cos(x) sin(x)). a) Show that f ′(x)/g′(x) converges to zero as x →∞. b) Verify that f(x)/g(x) remains in the interval [1/e, e] but does not converge. The function is not differentiable at ∞. There is no problem with l’Hopital. 4 Take the same functions from the previous example and look at the limit f(x)/g(x) for x →0. Now things are nice and dandy because the functions are differentiable at 0. 5 a) Assume a function f(x) satisfies f(0) = 0 and f ′(0) ̸= 0. Verify the following formula lim x→0 f(ax)/f(bx) = a/b . b) Given a differentiable function g satisfying g(0) = 0 and a differentiable function f. Verify that lim x→0 f(g(x)) g(x) = f ′(0) . Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 13: Worksheet Hopital’s rule 1 What does l’Hopital’s rule say about lim x→0 exp(2x) −1 x . -1.0 -0.5 0.5 1.0 -1 1 2 3 2 Apply l’Hopital’s rule to get the limit of f(x) = sin(100x) sin(200x) for x →0. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 14: Newton’s method In the intermediate value theorem lecture, we have seen a simple method to find a root of a function: start with an interval [a, b] such that f(a) < 0 and f(b) > 0, then successively half the interval always choosing the side on which the function takes different signs at the boundary. We are then (b −a)/2n close to the root in n steps. If the function is differentiable we can do much better and use the value of the derivative at a boundary point to get closer. If we draw a tangent at (x, f(x)), then f ′(x) = f(x) x −T(x) . because f ′(x) is the slope of the tangent and the right hand side is ”rise” over ”run”. If we solve for T(x) we get The Newton map is defined as T(x) = x −f(x) f ′(x) . Newton’s method is the process to apply this map again and again until we are sufficiently close to the root. It is an extremely fast method to find the root of a function. Start with a point x, then compute a new point x1 = T(x), where T(x) = x −f(x)/f ′(x) . Now iterate this again and again. If p is a root such that f ′(p) ̸= 0, and x0 is close to p, then x1 = T(x), x2 = T 2(x) converges to the root p. fHxL x-THxL x0 x1 x2 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0.5 1.0 1.5 1 If f(x) = ax + b, we reach the root in one step. 2 If f(x) = x2 then T(x) = x −x2/(2x) = x/2. We get exponentially fast to the root 0 but not as fast as the method promises. Indeed, the root is also a critical point which slows us down. 3 The Newton map brings us to infinity if we start at a critical point. Newton used the method to find the roots of polynomials. The method is so fast that it amazes: Starting 0.1 close to the point, we have after one step 0.01 after 2 steps 0.0001 after 3 steps 0.00000001 and after 4 steps 0.0000000000000001. 1 2 The Newton method converges extremely fast to a root f(p) = 0 if f ′(p) ̸= 0 if we start sufficiently close to the root. In 10 steps we can get a 210 = 1024 digits accuracy. Having a fast method to compute roots is useful. For example in computer graphics, where things can not be fast enough. Also in number theory, when working with integers having thousands of digits the Newton method can help. Besides that, there is theoretical use which can explain for example the stability of planetary motion. 4 Verify that the Newton map T(x) in the case f(x) = (x −1)3 has the property that we approach the root x = 1. Solution. You see that the approach is not that fast: we get T(x) = x+(1−x)/3 = (1+2x)/3. It converges exponentially fast, but not superexponential. T The reason is that the derivative at x −1 is not zero. That slows us down. If we have several roots, and we start at some point, to which root will the Newton method converge? Doe it at all converge? This is an interesting question. It is also histor-ically intriguing because it is one of the first cases, where ”chaos” can be observed at the end of the 19’th century. 5 Find the Newton map in the case f(x) = x5 −1. Solution T(x) = x −(x5 −1)/(5x4). If we look for roots in the complex like for f(x) = x5 −1 which has 5 roots in the complex plane, the basin of attraction of each of the points is a complicated set, a so called Newton fractal. Here is the picture: 3 6 Lets compute √ 2 to 12 digits accuracy - by hand! We want to find a root f(x) = x2 −2. The Newton map is T(x) = x −(x2 −2)/(2x). Lets start with x = 1. T(1) = 1 −(1 −2)/2 = 3/2 T(3/2) = 3/2 −((3/2)2 −2)/3 = 17/12 T(17/12) = 577/408 T(577/408) = 665857/470832 This is already 1.6 · 10−12 close to the real root! 7 To find the cube root of 10 we have to find a root of f(x) = x3 −10. The Newton map is T(x) = x −(x3 −10)/(3x2). If we start with x = 2, we get the following steps 2, 13/6, 3277/1521, 105569067476/49000820427. After three steps we have a result which is already 2.2 · 10−9 close to the root. The Newton method is an incredibly fast algorithm to get roots x0 of equations. Simply scrumtrulescent. 4 Homework 1 Find the Newton map T(x) = x −f(x)/f ′(x) in the following cases a) f(x) = x3 b) f(x) = ex c) f(x) = e−x2 d) f(x) = 2 tan(x). 2 a) The sinc function f(x) = sin(x)/x has a root between 1 and 4. We get closer to the root by doing a Newton step starting with x = π/2. Do this step -0.2 0.2 0.4 0.6 0.8 1.0 1.2 -0.5 0.5 1.0 3 The Newton map is handy to compute square roots. Assume we cant to find the square root of 99. We have to solve √ 99 = x or f(x) = x2 −99 = 0. Perform two Newton steps T(x) = x −(x2 −99)/(2x) starting at x = 10. 4 a) Find the Newton step T(x) = x−f(x)/f ′(x) in the case f(x) = 1/x and f(x) = x6. b) Find the Newton step T(x) in general if f(x) = xα, where α is a real number. 5 A chaotic Newton map. Verify that the Newton map in the case f(x) = (4 −3/x)1/3 is the quadratic map T(x) = 4x(1 −x). We will see a demonstration in class which shows that this map is a true random number generator. The Newton map does not converge. 0.2 0.4 0.6 0.8 1.0 -6 -4 -2 The graph of the function f(x) and a few Newton steps. The function is continuous on (0, 1). Its derivative too except at x = 2/3. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 14: Worksheet Newton Method 1 In the following graph, trace the two Newton steps already done. Add one more! x0 x1 x2 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0.5 1.0 1.5 2 In the following graph, try a few Newton steps. Let your start-ing point x0 be around 0.4. 0.2 0.4 0.6 0.8 1.0 1.2 1.4 -2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0 3 We will together compute the square root of 2 to an accuracy of 12 digits. Without computer. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 15: Review for first midterm Major points A function is continuous, if the closeness of x, y implies the closeness of f(x), f(y). Intermediate value theorem: f(a) > 0, f(b) < 0 implies f having a root in (a, b). At a local extremum, f ′(x) = 0. If f ′′(x) > 0, it is a local minimum. If f ′′(x) > 0, it is a local maximum. Global extrema: compare local extrema and boundary values. If f ′ > 0 then f is increasing, if f ′ < 0 it is decreasing. If f ′′(x) > 0 it is concave up, if f ′′(x) < 0 it is concave down. If f ′(x) = 0 then f has a horizontal tangent. Hoptial tells that limits limx→p f(x)/g(x), where f(p) = g(p) = 0 or f(p) = g(p) = ∞with g′(p) ̸= 0 are given by f ′(p)/g′(p). With Df(x) = (f(x + h) −f(x))/h and S(x) = h(f(h) + f(2h) + ...f(kh)) we have SDf(kh) = f(kh) −f(0) and DS(f(kh)) = f(kh). This is a preliminary fundamental theorem of calculus. Roots of f(x) with f(a) < 0, f(b) > 0 can be obtained numerically by dissection or by applying the Newton map T(x) = x −f(x)/f ′(x) again and again. Algebra reminders Healing: (a + b)(a −b) = a2 −b2 or 1 + a + a2 + a3 + a4 = (a5 −1)/(a −1) Denominator: 1/a + 1/b = (a + b)/(ab) Exponential: (ea)b = eab, eaeb = ea+b, ab = eb log(a) Logarithm: log(ab) = log(a) + log(b). log(ab) = b log(a) Trig functions: cos2(x) + sin2(x) = 1, sin(2x) = 2 sin(x) cos(x), cos(2x) = cos2(x) −sin2(x) Square roots: a1/2 = √a, a−1/2 = 1/√a Important functions Polynomials x3 + 2x2 + 3x + 1 Rational functions (x + 1)/(x3 + 2x + 1) Trig functions 2 cos(3x) Exponential 5e3x Logarithm log(3x) Inverse trig functions arctan(x) Important derivatives 1 2 f(x) f ′(x) f(x) = xn nxn−1 f(x) = eax aea·x f(x) = cos(ax) −a sin(a · x) f(x) f ′(x) f(x) = sin(ax) a cos(a · x) f(x) = tan(x) 1/ cos2(x) f(x) = log(x) 1/x Differentation rules Addition rule (f + g)′ = f ′ + g′. Scaling rule (cf)′ = cf ′. Product rule (fg)′ = f ′g + fg′. Quotient rule (f/g)′ = (f ′g −fg′)/g2. Chain rule (f(g(x))′ = f ′(g(x))g′(x). Easy rule simplify before deriving Extremal problems 1 Build a fence of length x+2y = 12 with dimensions x and y with maximal area A = xy. 2 Find the largest area A = 4xy of a rectangle with vertices (x, y), (−x, y), (−x, −y), (x, −y) inscribed in the ellipse x2 + 2y2 = 1. 3 Which isosceles triangle of height h and base 2x and area xh = 1 has minimal circumference 2x + 2 √ x2 + h2? 4 Where is the distance p x2 + y2 of the parabola y = x2 = 2 to the point (0, 0) minimal? 5 A cone of height h = 1 + x and radius r = √ 1 −x2 is tightly enclosed by a unit sphere centered at height x. Maximize the volume r2πh/3 of the cone. 6 Maximize f(x) = sin(x) on [0, π]. Limit examples limx→0 sin(x)/x l’Hopital 0/0 limx→0(1 −cos(x))/x2 l’Hopital 0/0 twice limx→0 x log(x) l’Hopital ∞/∞ limx→1(x2 −1)/(x + 1) heal directly limx→∞exp(x)/(1 + exp(x)) l’Hopital limx→0(x + 1)/(x + 5) no work necessary Important things Summation and taking differences is at the hart of calculus The 3 major types of discontinuities are jump, oscillation, infinity The Newton method is an algorithm to find roots Remember the fundamental theorem of trigonometry limx→0 sin(x)/x = 1. The derivative is the limit Df(x) = [f(x + h) −f(x)]/h as h →0. It is called rate of change. The rule D(1 + h)x/h = (1 + h)x/h leads to exp′(x) = exp(x). More Examples 1 Find limx→1(x1/4 −1)/(x1/5 −1). Answer: 5/4. 2 Find limx→1 sin(4x −4)/(x −1). Answer: 4. 3 Find limx→2 3−√7+x x−2 . Answer −1/6 4 Find arcsin′(5x2). Answer: 10x(1 −25x4)−1/2 5 Is 1/ log |x| continuous at x = 0. Answer: yes 6 Is log(1/|x|) continuous at x = 0. Answer: no Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 15: Worksheet Checklist Make a list of the most important definitions and a list of the most important results in this course. Mind map Produce your own mind map of the course. Here are some starting points. On the back is a suggestion. On the abac Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 16: Mean value theorem In this lecture, we look at the mean value theorem and a special case called Rolle’s theorem. It is important later when we study the fundamental theorem of calculus. Unlike the intermediate value theorem which applied for continuous functions, the mean value theorem involves derivatives: Mean value theorem: For a differentiable function f and an interval (a, b), there exists a point p inside the interval such that f ′(p) = f(b) −f(a) b −a . fHbL-fHaL b-a Here are a few examples which illustrate this: 1 The function f(x) = x2 + ax + b has roots at u = (−a ± √ a2 −4b)/2. The derivative 2x + a = 0 is zero for x = a/2. 2 f(x) = cx, then f ′(x) = cx and f(b) −f(a) = (cb −ca)/(b −a) = c. So, every point x has the derivative c. 3 f(x) = arcsin(x) has the property that for any x, y in (−1, 1), we have | arcsin(y) − arcsin(x)| ≥|x −y|. Solution. The derivative of arcsin(x) is 1/ √ 1 −x2 > 1. -1.5 -1.0 -0.5 0.5 1.0 1.5 -1.5 -1.0 -0.5 0.5 1.0 1.5 -1 1 2 3 -1 1 2 3 4 A biker drives with velocity f ′(t) at position f(b) at time b and at position a at time a. The value f(b) −f(a) is the distance traveled. The fraction [f(b) −f(a)]/(b −a) is the 1 2 average speed. The theorem tells that there was a time when the bike had exactly the average speed. 5 The function f(x) = √ 1 −x2 has a graph on (−1, 1) on which every possible slope is taken. Solution: We can see this with the intermediate value theorem because f ′(x) = x/ √ 1 −x2 gets arbitrary large near x = −1 or x = 1. The mean value theorem shows this too because we can take intervals [a, b] = [−1, −1 + c] for which [f(b) −f(a)]/(b −a) = f(−1 + c)/ci ∼ √c/c = 1/√c gets arbitrary large. Why is the theorem true? The function h(x) = f(a) + cx, where c = (f(b) −f(a)/(b −a) also connects the beginning and end point. The function g(x) = f(x)−h(x) has now the property that g(a) = g(b). If we can show that for such a function, there exists x with g′(x) = 0, then we are done. By tilting the picture, we have reduced the statement to a special case which is important by itself: Rolle’s theorem: If f(a) = f(b) and f is differentiable, then there exists a critical point p of f in the interval (a, b). Here is the proof: If it were not true, then f ′(x) ̸= 0 and we would have f ′(x) > 0 everywhere or f ′(x) < 0 everywhere. This would mean however that f(b) > f(a) or f(b) < f(a). Here is a second proof: Fermat’s theorem assures that there is a local maximum or local minimum of f in (a, b). At this point the derivative is zero. This means f ′(x) = 0. We have also seen a related fact that if f is continuous and f(a) = f(b) then there is a local maximum or local minimum in the interval (a, b). This fact is more general and applies to every continuous function. The derivative does not need to exist. 6 There is a point in [0, 1] where f ′(x) = 0 with f(x) = x(1 −x2)(1 −sin(πx)). Solution: We have f(0) = f(1) = 0. Use Rolle. 7 Show that the function f(x) = sin(x) + x(π −x) has a critical point [0, π]. Solution: The function is nonnegative and zero at 0, π.It is also differentiable and so by Rolle’s theorem there is a critical point. Remark. We can not use Rolles theorem to show that there is a local maximum even so the extremal value theorem assures us that this exist. 8 Verify that the function f(x) = 2x3 + 3x2 + 6x + 1 has only one real root. Solution: There is one real root by the intermediate value theorem: f(−1) = −4, f(1) = 12. Assume 3 there would be two roots. Then by Rolles theorem there would be a value x where g(x) = f ′(x) = 6x2 + 6x + 6 has a root. But there is no root of g. [The graph of g minimum at g′(x) = 6 + 12x = 0 which is 1/2 where g(1/2) = 21/2 > 0.] Who was the first to find the mean value theorem? It is not so clear. Joseph Louis Lagrange is one candidate. Also Augustin Louis Cauchy (1789-1857) is credited for a modern formulation of the theorem. Joseph Louis Lagrange, 1736-1813. Augustin Louis Cauchy, 1789-1857. What about Michel Rolle? He lived from 1652 to 1719, mostly in Paris. No picture of him seems available. Rolle also introduced the n’th root notation like 3 √x. 4 Homework 1 The function f(x) = 1 −|x| satisfies f(−1) = f(1) = 0 but there is no point where f ′(x) = 0. Is this a counter example to Rolle’s theorem? 2 Use Rolles theorem and the intermediate value theorem to show that the function f(x) = x3 + 3x + 1 has exactly one root. You do not have to find the root. 3 We look at the function f(x) = log |x| + sin(x) on the positive real line Use the intermediate value theorem applied to f ′(x) to assure that for every M > 0 there is a positive x for which f ′(x) = M. Use the the mean value theorem to assure that we can find for every M two values a, b such that f(b) −f(a)/(b −a) = M. 4 Cauchy’s mean value theorem states that for any two differentiable function and any interval (a, b), there exists c for which (f(b) −f(a))g′(c) = (g(b) −g(a))f ′(c). To prove this , define the function h(x) = (f(b) −f(a))(g(x) −g(a)) −(g(b) −g(a))(f(x) −f(a)). a) Verify that h(a) = h(b) = 0. b) Compute h′(x). c) Use Rolle’s theorem to verify that there is a c for which h′(c) = 0. Hint. If stuck, there is more explanation in http : //en.wikipedia.org/wiki/Mean value theorem. But first give it a shot on your own. 5 Given the function f(x) = x sin(x) and the function g(x) = cos(x). Verify (using Cauchy’s mean value theorem) that there is a point p ∈(0, π/2) for which f ′(p)/g′(p) = −π/2. You do not have to find the point. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 16: Worksheet The mean value theorem In this class, we have at various places looked at calculus with discrete eyes, where Df(x) = [f(x + h) −f(x)]/h . We look here at the question whether there is a dis-crete version of Rolle’s theorem. You may have the opportunity to find a new result. Note that quantum results hold in general for functions which are only continuous. No differentiability is needed. This worksheet might give you an idea what re-search is about. You do not need the answer yet, whether a result works or not. It is exciting because nobody else does now simply because nobody has studied the question yet! 1 Quantum Rolle: Given an interval [a, b] from which we assume that its length is larger than h. Given a continuous function f such that f(a) = f(b) = 0. Is it true that there is a point p in that interval for which Df(p) = 0? Play and doodle around with examples. 2 Quantum mean value theorem: Given an interval [a, b] and a function f. Is it true that there is a point p such that Df(p) = f(b) −f(a) b −a ? Play around with examples. 3 Argue that you can ”tilt” the setting as in the continuum so that if the quantum Rolle re-sult holds, then the quantum mean value theorem holds. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 17: Catastrophes In this lecture, we once more cover extrema problems. We are interested how extrema change when a parameter changes. Nature, economies, processes favor extrema. Extrema change smoothly with parameters. How come that the outcome is often not smooth? What is the reason that political change can go so fast once a tipping point is reached? One can explain this with mathemati-cal models. We look at a simple example, which explains it. In reality, the situation is more complicated. In the ”New York Times” of February 24, 2011, Jennifer E. Sims, the director of intelligence studies at Georgetown University’s School of Foreign Service and senior fellow at the Chicago Council on Global Affairs asked: Why, with the U.S. spending 80 billion dollars on intelligence, were we apparently surprised by recent regime changes in the Middle East? Why did change happen at all? These are complex questions. Obviously, some tipping point has been reached and the smallest event like the confiscation of a fruit stand in Tunesia or increasing food prizes in Egypt has produced change. In these complex examples, it will never be possible to un-derstand everything. Lets look here at a simple mathematical model which illustrates the general principle that: If a local minimum seizes to become a local minimum, a new stable position is favored. This can be far away from the original situation. To get started, lets look at an extremization problem 1 Find all the extrema of the function f(x) = x4−x2. So-lution: f ′(x) = 4x3−2x is zero for x = 0, 1/ √ 2, −1/ √ 2. The second derivative is 12x2−2. It is negative for x = 0 and positive at the other two points. We have two local minima and one local maximum. x4 - x2 x 2 Now find all the extrema of the function f(x) = x4 − x2 −2x. There is only one critical point. It is x = 1. x4 - x2 - 2 x x Something has happened from the first example to the second example. The local minimum to the left has disappeared. Assume the function f measures the prosperity of some kind and c is 1 2 a parameter. We look at the position of the first equilibrium point of the function. Catastroph theorists usually assume the so called Delay assumption. A stable equilibrium is here used as an other name for a local minimum. A system state remains in a stable equilibrium until it disappears. If that happens, the system settles in a neighboring stable equilibrium. A parameter value for which a stable minimum disappears is called a catastrophe. Here is the position of the equilibrium point plotted in dependence of c. 3 c f A parameter value for which a local minimum disappears is called a catastrophe. c Bifurcation diagram: The picture shows the equilibrium points as they change in dependence of the parameter c. The vertical axes is the parameter c, the horizontal axes is x. At the bottom for c = 0, we have three equilibrium points, two local minima and one local maximum. At the top for c = 1 we have only one local minimum. Catastrophes always go for the worse in the sense that the value decreases. It is not possible to reverse the process and have a catastrophe, where the minimum jumps up. Look again at the above ”movie” of graphs. But run it backwards and use the same principle. We do not end up at the position we started with. The new equilibrium stays the equilibrium. Decreasing the food prizes again did not reverse the process of change in Egypt for example. Catastrophes are in general irreversible. We see that in real life: It is easy to screw up a relationship, get sick, have a ligament torn or lose trust. Building up a relationship, getting healthy or gaining trust on the other hand happens 4 slowly. Ruining a country or a company or losing a good reputation of a brand is very easy. It takes a long time to regain it. Local minima can change discontinuously, when a parameter is changed. This can happen with perfectly smooth functions and smooth parameter changes. 3 Lets look at f(x) = x4 + cx2, where −1 ≤c ≤1. We will look at that in class. c Homework In this homework, we study a catastrophe for the function f(x) = x6 −x4 + cx2 , where c is a parameter between 0 and 1. 1 a) Find all the critical points in the case c = 0 and analyze their stability. b) Find all the critical points in the case c = 1 and analyze their stability. 2 Plot the graph of f for at least 10 values of c between 0 and 1. You can of course use software, a graphing calculator or Wolfram alpha. Mathematica code is below. 3 If you change from c = −0.3 to 0.6 pinpoint the value for the catastrophe and show a rough plot of c →f(xc), the value at the first local minimum xc in dependence of c. The text above provides this graph for an other function. It is the graph with a discontinuity. 4 If you change back from c = 0.6 to 0.3 pinpoint the value for the catastrophe (it will be different from the one in the previous question). 5 Sketch the bifurcation diagram. That is, if xk(c) is the k’th equilibrium point, then draw the union of all graphs of xk(c) as a function of c (the c-axes pointing upwards). As in the two example provided, draw the local maximum with dotted lines. ✞ Manipulate [ Plot [ xˆ6 −xˆ4 + c xˆ2 , {x , −1, 1}] , {c , 0 , 1}] ✝ ✆ Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 17: Worksheet Catastrophes We see here graphs of the function f(x) = x4 −cx2 for c between 0 and 1: 1 Draw the bifurcation diagram in this case. The vertical axes is the c axes. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 18: Riemann integral In this lecture we define the integral R x 0 f(t) dt if f is a differentiable function and compute it for some basic functions. First a reminder. We have defined the Riemann sums Sf(x) = h[ f(0) + f(h) + f(2h) + .... + f(kh)] , where k is the largest integer such that kh < x. Lets write Sn if we want to stress that the parameter h = 1/n was used in the sum. We define the integral as the limit of these sums Snf when the mesh size h = 1/n goes to zero. Define Z x 0 f(t) dt = lim n→0 Snf(x) . xk x y 0 For any differentiable function, the limit exists Proof: Lets first assume f ≥0 on [0, x]. Let M be such that f ≤M and f ′ ≤M on [0, x]. The Riemann sum Snf(x) is the total area of k rectangles. Let Sf(x) denote the area under the curve. If M is the maximal slope of f on [0, x], then on each interval [j/n, (j + 1)/n] , we have |f(x) −f(j/n)| ≤M/n so that the area error is smaller than M/n2. Additionally, we have a piece above the interval [kh, x] with area ≤M/n. If we add all the k ≤xn ”roof area errors” and the ”side area” up, we get Sf(x) −Snf(x) ≤kM n2 + M n ≤xnM n2 + M n = xM + M n . This converges to 0 for n →∞. The limit is therefore the area Sf(x). For a general, not necessarily nonnegative function, we write f = g −h, where g, h are nonnegative (see homework) and have R x 0 f(x) dx = R x 0 g(x) dx − R x 0 h(x) dx. For nonnegative f, the value R x 0 f(x) dx is the area between the x-axis and the graph of f. For general f, it is a signed area, the difference between two areas. Remark: the Riemann integral is defined here as the limit h P xk=kh∈[0,x) f(xk). It converges to the area under the curve for all continuous functions but since we work with differentiable functions in calculus we restricted to that. Not all bounded functions can be integrated naturally 1 2 like this. There are discontinuous functions like the salt and pepper function which is defined to be f(x) = 1 if x is rational and zero else. Now Sf(x) = 1 for rational h and Sf(x) = 0 if h is irrational. Therefore, an other integral, the Lebesgue integral is used too: it can be defined as the limit 1 n Pn k=1 f(xk) where xk are random points in [0, x]. This Monte-Carlo integral definition of the Lebesgue integral gives the integral 0 for the salt and pepper function because rational numbers have zero probability. Remark: Many calculus books define the Riemann integral using partitions x0 < x1 < ..., xn of points of the interval [0, x] such that the maximal distance (xk+1 −xk) between neighboring xj goes to zero. The Riemann sum is then Snf = P k f(yk)(xk+1 −xk), where yk is arbitrarily chosen inside the interval (xk, xk+1). For continuous functions, the limiting result is the same the Sf(x) sum done here. There are numerical reasons to allow more general partitions because it allows to adapt the mesh size: use more points where the function is complicated and keep a wide mesh, where the function does not change much. This leads to numerical analysis of integrals. 1 Let f(x) = c be constant everywhere. Now R x 0 f(t) dt = cx. We can see also that cnx/n ≤Snf(x) ≤c(n + 1)x/n. 2 Let f(x) = cx. The area is half of a rectangle of width x and height cx so that the area is cx2/2. Remark: we could also have added up the Riemann sum but thats more painful: for every h = 1/n, let k be the largest integer smaller than xn = x/h. Then (remember Gauss’s punishment?) Snf(x) = 1 n k X j=1 cj n = ck(k + 1)/2 n2 . Taking the limit n →∞and using that k/n →x shows that R x 0 f(t) dt = cx2/2. 3 Let f(x) = x2. In this case,we can not see the numerical value of the area geometrically. But since we have computed S[x2] in the first lecture of this course and seen that it is [x3]/3 and since we have defined Shf(x) → R x 0 f(t) dt for h →0 and [xk] →xk for h →0, we know that Z x 0 t2 dt = x3 3 . 3 This example actually computes the volume of a pyramid which has at distance t from the top an area t2 cross section. Think about t2dt as a slice of the pyramid of area t2 and height dt. Adding up the volums of all these slices gives the volume. Linearity of the integral (see homework) R x 0 f(t)+g(t) dt = R t 0 f(t) dt+ R x 0 g(t) dt and R x 0 λf(t) dt = λ R x 0 f(t) dt. Upper bound: If 0 ≤f(x) ≤M for all x, then R x 0 f(t) dt ≤Mx. 4 R x 0 sin2(sin(sin(t)))/x dt ≤x. Solution. The function f(t) inside the interval is nonnega-tive and smaller or equal to 1 The graph of f is therefore contained in a rectangle of width x and height 1. We see that if two functions are close then their difference is a function which is included in a small rectangle and therefore has a small integral: If f and g satisfy |f(x) −g(x)| ≤c, then Z x 0 |f(x) −g(x)| dx ≤cx . We know identities like Sn[x]n h = [x]n+1 h n+1 and Sn exph(x) = exph(x) already. Since [x]k h − [x]k →0 we have Sn[x]k h −Sn[x]k →0 and from Sn[x]k h = [x]k+1 h /(k + 1). The other equalities are the same since exph(x) = exp(x) →0. This gives us: R x 0 tn dt = xn+1 n+1 R x 0 et dt = ex −1 R x 0 cos(t) dt = sin(x) R x 0 sin(t) dt = 1 −cos(x) 4 Homework 1 a) Find the integral R x 0 t5 + 4t3 + et dt. b) Calculate R 10 0 t3 −t + t2 dt. c) Find R 3π −5π cos(t) dt. 2 Verify that the following statements hold for differentiable functions f, g and a < b < c and any real number λ. You can argue geometrically with areas. • R b a f(x) dx + R c b f(x) dx = R c a f(x) dx. • R b a f(x) dx + R b a g(x) dx = R b a f(x) + g(x) dx. • R b a λf(x) dx = λ R b a f(x) dx. • 0 ≤m ≤f(x) ≤M implies (b −a)m ≤ R b a f(x) dx ≤(b −a)M. 3 a) Verify that every differentiable function f can be written as a difference of two non-negative functions. To do so, show that g(x) = max(f(x), 0) and h(x) = max(−f(x), 0) have the property that f(x) = g(x) −h(x) and that g(x) ≥0 and h(x) ≥0. b) Draw the graphs of the two functions g(x), h(x) in the case f(x) = sin(3x) where 0 ≤x ≤2π. 4 a) The region enclosed by the graph of x and the graph of x3 has a propeller type shape as seen in the picture. Find its (positive) area. b) What is the integral R 2π 0 | sin(x)| dx? 5 a) Find R 3 0 |x −1| dx. Distinguish cases. b) Find R 3 0 f(x) dx for f(x) = |x −|x −1|| · |x −2|. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 18: Worksheet Riemann sums We look at the function f(x) = exp(−x2). It is a function for which the integral R x 0 f(t) dt is not elementary. We can not express it with polynomials, trig, exponential functions or their inverses. We want here to get estimates for R 1 0 exp(−x2) dt. 1 1 2 1 3 1 Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 19: Fundamental theorem In this lecture we prove the fundamental theorem of calculus for differentiable functions. This will allow us in general to compute integrals of functions which appear as derivatives. We have seen earlier that with Sf(x) = h(f(0) + · · · + f(kh)) and Df(x) = (f(x + h) −f(x))/h we have SDf = f(x) −f(0) and DSf(x) = f(x) if x = nh. This becomes now: Fundamental theorem of calculus: Assume f is differentiable. Then R x 0 f ′(t) dt = f(x) −f(0) and d dx R x 0 f(t) dt = f(x) Proof. Using notation of Euler we write A ∼B for ”A and B are close” meaning A −B →0 for h →0. From DSf(x) = f(x) for x = kh we have DSf(x) ∼f(x) for kh < x < (k + 1)h because f is continuous. We also know R x 0 Df(t) dt ∼ R x 0 f ′(t) dt because Df(t) ∼f ′(t) uniformly for 0 ≤t ≤x by the definition of the derivative and the assumption that f ′ is continuous. We also know SDf(x) = f(x)−f(0) for x = kh. By definition of the Riemann integral Sf(x) ∼ R x 0 f(t) dt and so SDf(x) ∼ R x 0 Df(t) dt. f(x) −f(0) ∼SDf(x) ∼ Z x 0 Df(t) dt ∼ Z x 0 f ′(t) dt as well as f(x) ∼DSf(x) ∼D Z x 0 f(t) dt ∼d dx Z x 0 f(t) dt . 1 R 5 0 3t7 dt = x8 8 |5 0 = 58 8 . You can always leave such expressions as your final result. It is even more elegant than the actual number 390625/8. 2 R π/2 0 cos(t) dt = sin(x)|π/2 0 = 1. This is an important example which should become routine in a while. 3 R x 0 √1 + t dt = R x 0 (1+t)1/2 dt = (1+t)3/2(2/3)|x 0 = (1+x)3/2 −1. Here the difficulty was to see that the 1 + t in the interior of the function does not make a big difference. Keep such examples in mind. 4 Also in this example R 2 0 cos(t + 1) dt = sin(x + 1)|2 0 = sin(3) −sin(1) the additional term +1 does not make a big dent. 5 R π/4 π/6 cot(x) dx. This is an example where the anti derivative is difficult to spot. It is easy if we know where to look for: the function log(sin(x)) has the derivative cos(x)/ sin(x). So, we know the answer is log(sin(x))|π/4 π/6 = log(sin(π/4))−log(sin(π/6)) = log(1/ √ 2)−log(1/2) = −log(2)/2 + log(2) = log(2)/2. 6 The example R 2 1 1/(t2 −9) dt is a bit challenging. We need a hint and write −6/(x2 −9) = 1/(x+3)−1/(x−3). The function f(x) = log |x+3|−log |x−3| has therefore −6/(x2 −9) as a derivative. We know therefore R 2 1 −6/(t2 −9) dt = log |3 + x| −log |3 −x||2 1 = log(5) −log(1) −log(4) + log(2) = log(5/2). The original task is now (−1/6) log(5/2). 7 R x 0 cos(sin(x)) cos(x) dx = sin(sin(x)) because the derivative of sin(sin(x)) is cos(sin(x)) cos(x). The function sin(sin(x)) is called the antiderivative of f. If we differentiate this function, we get cos(sin(x)) cos(x). 8 Find R π 0 sin(x) dx. Solution: This has a very nice answer. 1 2 Here is an important notation, which we have used in the example and which might at first look silly. But it is a handy intermediate step when doing the computation. F|b a = F(b) −F(a). We give reformulations of the fundamental theorem in ways in which it is mostly used: If f is the derivative of a function F then Z b a f(x) dx = F(x)|b a = F(b) −F(a) . In some textbooks, this is called the ”second fundamental theorem” or the ”evaluation part” of the fundamental theorem of calculus. The statement d dx R x 0 f(t) dt = f(x) is the ”antiderivative part” of the fundamental theorem. They obviously belong together and are two different sides of the same coin. Here is a version of the fundamental theorem, where the boundaries are functions of x. Given functions g, h and if F is a function such that F ′ = f, then Z g(x) h(x) f(t) dt = F(g(x)) −F(h(x)) . 9 R x2 x4 cos(t) dt = sin(x2) −sin(x4). The function F is called an antiderivative. It is not unique but the above formula does always give the right result. Lets look at a list of important antiderivatives. You should have as many antideriatives ”hard wired” in your brain. It really helps. Here are the core functions you should know. They appear a lot. function anti derivative xn xn+1 n+1 √x x3/22 3 eax eax a cos(ax) sin(ax) a sin(ax) −cos(ax) a 1 x log(x) 1 1+x2 arctan(x) log(x) x log(x) −x Make your own table! 3 Meet Isaac Newton and Gottfried Leibniz. They have discovered the fundamental theorem of calculus. You can see from the expression of their faces how honored they are to find themselves on the same handout with Austin Powers and Doctor Evil. Culture clash ... 4 Homework 1 For any of the following functions f, find a function F such that F ′ = f. a) ex + sin(3x) + x3 + 5x. b) (x + 4)3. c) 1/x + 1/(x −1). d) cos(x2)2x + sin(x3)3x2 + 1/√x 2 Find the following integrals by finding a function g satisfying g′ = f. We will learn techniques to find the function. Here, we just use our knowledge about derivatives: a) R 3 2 5x4 + 4x3 dx. b) R π/2 π/4 sin(3x) + cos(x) dx. c) R π/2 π/4 1 sin2(x) dx. d) R 3 2 1 x−1 dx. ‘ 3 Evaluate the following integrals: a) R 2 1 2x dx. b) R 1 −1 cosh(x) dx. (Remember cosh(x) = (ex + e−x)/2.) c) R 1 0 1 1+x2 dx. d) R 2/3 1/3 1 √ 1−x2 dx. 4 a) Compute F(x) = R x3 0 sin(t) dt, then find F ′(x). b) Compute G(x) = R cos(x) sin(x) exp(t) dt then find G′(x) 5 a) Be clever: Evaluate the following integral: R 2π 0 sin(sin(x)) dx Give the answer and the reason in a short sentence. b) Be evil: Take a function F of your choice. Find its derivative and call it f. Now pose an integration problem to find R b a f(x) dx. Submit this problem to knill@math.harvard.edu I will select the most evil one. A good problem should lead to a short function f but the integral F should be difficult to find or guess. These problems will make perfect exam problems for the sec-ond midterm .... (evil laugh). You can submit your version of Problem 5b) electronically by email (knill@math.harvard.edu. Just send the function in the subject line. Mail can otherwise be empty). For any submission, independent how clever or evil, 10 points maximal will be added to your score (maxing up at 50). So, if your HW score of Lecture 19 is 45 and you submitted a function, it will be bumped to 50. If your HW score is 50 already you get nothing ... (more evil laugh). Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 19: Worksheet Fundamental theorem Find the following integrals 1 R 3 1 x3 dx 2 R 1 −2 1 −x5 dx 3 R 1 0 1 x2+1 dx 4 R 4 2 1 x dx Now a bit harder ones: 5 R 4 1 x1/3 dx 6 R 3 1 √ 1 + x5 dx 7 R 2 1 1 √x + 5 x2 dx 8 R 2 1 1 1+x5 dx Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 20: Antiderivatives We have looked at the integral R x 0 f(t) dt and seen that it is the signed area under the curve. We have seen that the area of the region below the curve is counted in a negative way. There is something else to mention: For x < 0, we define R x 0 f(t) dt as − R 0 x f(t) dt. This is compatible with the funda-mental theorem R b a f ′(t) dt = f(b) −f(a). We call g(x) = R x 0 f(t) dt + C an anti-derivative of g. The constant C is arbitrary and not fixed. As we will see below, we can often adjust the constant such that some condition is satisfied. The fundamental theorem of calculus assured us that The antiderivative is the inverse operation of the derivative. Two different anti derivatives differ by a constant. Finding the anti-derivative of a function is much harder than finding the derivative. We will learn some techniques but it is in general not possible to give anti derivatives for even very simple functions. 1 Find the anti-derivative of f(x) = sin(4x) + 20x3 + 1/x. Solution: We can take the anti-derivative of each term separately. The antiderivative is F(x) = −cos(4x)/4 + 4x4 + log(x) + C. 2 Find the anti derivative of f(x) = 1/ cos2(x) + 1/(1 −x). Solution: we can find the anti derivatives of each term separately and add them up. The result is F(x) = cot(x)+log |1− x| + C. 3 We mentioned Galileo Galileo, who measured free fall motion with constant acceleration. Assume s(t) is the position of the ball at time t. Assume the ball has zero velocity initially and is located at height s(0) = 20. We know that the velocity is v(t) is the derivative of s(t) and the acceleration a(t) is constant equal to −10. So, v(t) = −10t + C is the antiderivative of a. By looking at v at time t = 0 we see that C = v(0) is the initial velocity and so zero. We know now v(t) = −10t. We need now to compute the anti derivative of v(t). This is s(t) = −10t2/2 + C. Comparing t = 0 shows C = 20. Now s(t) = 20−5t2. The graph of s is a parabola. If we give the ball an additional horizontal velocity, such that time t is equal to x then s(x) = 20 −5x2 is the visible trajectory. We see that jumping from 20 meters leads to a fall which lasts 2 seconds. 1 2 0.5 1.0 1.5 2.0 5 10 15 20 4 The total cost is the antiderivative of the marginal cost of a good. Both the marginal cost as well as the total cost are a function of the quantity produced. For instance, suppose the total cost of making x shoes is 300 and the total cost of making x+4 shoes is 360 for all x. The marginal cost is 60/4 = 15 dollars. In general the marginal cost changes with the number of goods. There is additional cost needed to produce one more shoe if 300 shoes are produced. Problem: Assume the marginal cost of a book is f(x) = 5 −x/100 and that producing the first 10 books costs 1000 dollars. What is the total cost of producing 100 books? Answer: The anti derivative 5 −x/100 of f is F(x) = 5x −x2/100 + C where C is a constant. By comparing F(10) = 1000 we get 50 −100/100 + C = 1000 and so C = 951. the result is 951+5∗100−10′000/100 = 1351. The average book prize has gone down from 100 to 13.51 dollars. A function f is called elementary, if it can be constructed using addi-tion, subtraction, multiplication, division, compositions from polynomials or roots. In other words, an elementary function is built up with functions like x3, √·, exp, log, sin, cos, tan and arcsin, arccos, arctan. 5 The function f(x) = sin(sin(π+√x+x2))+log(1+exp((x6 +1)/(x2 +1))+(arctan(ex))1/3 is an elementary function. 6 The anti derivative of the sinc function is called the sine-integral Si(x) = Z x 0 sin(t) t dt . The function Si(x) is not an elementary function. 3 2 4 6 8 10 1.2 1.4 1.6 1.8 7 The offset logarithmic integral is defined as Li(x) = Z x 2 dt log(t) It is a specific anti-derivative. It is a good approximation of the number of prime numbers less than x. The graph below illustrates this. The second stair graph shows the number π(x) of primes below x. For example, π(10) = 4 because 2, 3, 5, 7 are the only primes below it. The function Li is not an elementary function. 10 15 20 2 4 6 8 8 The error function erf(x) = 2 √π Z x 0 e−t2 dt is important in statistics. It is not an elementary function. 4 0.5 1.0 1.5 2.0 0.2 0.4 0.6 0.8 1.0 The Mathematica command ”Integrate” uses about 500 pages of Mathematica code and 600 pages of C code. 1 Before software was doing this, tables of integrals like Gradshteyn and Ryzhik’s work were used. This 1200 page book is still useful and contains some integrals, which computer algebra systems have trouble with. Numerical evaluation What do we do when we have can not find the integral analytically? We can still compute it numerically. Here is an example: the function sin(sin(x)) also does not have an elementary anti-derivative. But we can compute the integral numerically with a computer algebra system like Mathematica: ✞ NIntegrate [ Sin [ Sin [ x ] ] , { x , 0 , 1 0 } ] ✝ ✆ Pillow problems We do not assign homework over spring break. If you have time, here are some integration riddles. We will learn techniques to deal with them. If you can not crack them, no problem. Maybe pick one or two and keep thinking about it over spring break. They make also good pillow problems, problems to think about while falling asleep. Try it. Sometimes, you might know the answer in the morning. Maybe you can guess a function which has f(x) as a derivative. 1 f(x) = log(x)/x. 2 f(x) = 1 x4−1. 3 f(x) = tan2(x). 4 f(x) = cos4(x). 5 f(x) = 1 x log(x). 1 Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 20: Worksheet Anti derivatives Here are some trickier anti derivative puzzles. We still have no inte-gration techniques and must rely on intuition and experiments to find the derivatives. It is often a puzzle because we can try to combine derivatives of known functions to get the given function. 1 Find the anti-derivative of the function f(x) = 1 + x 1 −x Hint. First compute the anti derivative of g(x) = 1 1 −x . Can you combine f and g in some way to make it fit? 2 Find the anti derivative of the function f(x) = sin(x3)3x2 . Hint. Think about the chain rule. 3 Find the anti-derivative of the function f(x) = sin(sin(x)) cos(x) . Hint. Think about the chain rule. 4 Find the anti-derivative of the function f(x) = 2x sin(x) + x2 cos(x) . Hint. Think about the product rule. 5 Find the anti-derivative of the function f(x) = eeeeeex · eeeeex · eeeex · eeex · eex · ex . Hint. There is no hint. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 21: Area computation If f(x) ≥0, then R b a f(x) dx is the area under the graph of f(x) and above the interval [a, b] on the x axes. As you have seen in a homework, any function can be written as f(x) = f +(x) −f −(x), where f +(x) ≥0 and f −(x) ≥0. This means that we can write any integral R b a f(x) dx as the difference of the area above the graph minus the area below the graph. Z b a f(x) dx = Z b a f +(x) −f −(x) dx . Here is the most common situation: If a region is enclosed by two graphs f ≤g and x is also enclosed between a and b then its area is R b a g(x) −f(x) dx. 1 Find the area of the region enclosed by the x-axes, the y-axes and the graph of the cos function. Solution: R π/2 0 cos(x) dx = 1. 2 Find the area of the region enclosed by the graphs f(x) = x2 and f(x) = x4. 3 Find the area of the region enclosed by the graphs f(x) = 1 −x2 and g(x) = x4. 1 2 4 Find the area of the region enclosed by a half circle of radius 1. Solution: The half circle is the graph of the function f(x) = √ 1 −x2. The area under the graph is Z 1 −1 √ 1 −x2 dx . Finding the anti-derivative is not so easy. We will find techniques to do so, for now we pop it together: we know that arcsin(x) has the derivative 1/ √ 1 −x2 and x √ 1 −x2 has the derivative √ 1 −x2 −x2/ √ 1 −x2. The sum of these two functions has the derivative √ 1 −x2 −(1 −x2)/ √ 1 −x2 = 2 √ 1 −x2. We find the anti derivative to be (x √ 1 −x2 + arcsin(x))/2. The area is therefore x √ 1 −x2 + arcsin(x) 2 |1 −1 = π 2 . 5 Find the area of the region between the graphs of f(x) = 1−|x|1/4 and g(x) = −1+|x|1/4. 3 6 Find the area under the curve of f(x) = 1/x2 between −6 and 6. Solution. R 6 −6 x−2 dx = −x−1|6 −6 = −1/6 −1/6 = −1/3. There is something fishy with this computation because f(x) is nonnegative so that the area should be positive. But we obtained a negative answer. What is going on? 7 Find the area between the curves x = 0 and x = 2 + sin(y), y = 2π and y = 0. Solution: We turn the picture by 90 degrees so that we compute the area under the curve y = 0, y = 2 + sin(x) and x = 2π and x = 0. 4 8 The grass problem. Find the area between the curves |x|1/3 and |x|1/2. Solution. This example illustrates how important it is to have a picture. This is good advise for any ”word problem” in mathematics. Use a picture of the situation while doing the computation. Homework 1 Find the area of the region enclosed by the graphs f(x) = x3 and g(x) = p |x|. 2 Find the area of the region enclosed by the four lines y = x, y = 3 −x, x = 1. 3 Find the area of the region enclosed by the curves y = 4π, y = 2π, x = −3 + sin(3y), x = 2 + sin(2y). 4 Write down an integral which gives the area of the elliptical region 4x4 + y2 ≤1. Evaluate the integral numerically using Wolfram alpha, Mathematica or any other software. 5 The graphs sin(x) and cos(x) −1 intersect at x = 0, 2π and a point between. They define a humming bird region,consisting of a larger region and a tail region. Find the area of each and assume the bird has its eye closed. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 21: Worksheet Area Computation In this worksheet we look at other regions. In order to find the area we have to turn our heads. 1 Lets compute the area of the region enclosed by the lines x = 0, x = √y, y = 0 and y = 4. In order to solve such an area problem, we have to draw a picture. We started doing that. Find ways to find the area. 2 Lets compute the area of the region enclosed by the lines x = 0, x = y2, y = 0 and y = 2. Now its your turn to draw a picture and compute the area. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 22: Volume computation To compute the volume of a solid, we cut it into slices perpendicularily along a line x. If A(x) is the area of the slice and the body is enclosed between a and b then V = R b a A(x) dx is the volume. Think of A(x)dx as the volume of a slice. The integral adds them up. 1 Compute the volume of a pyramid with square base length 2 and height 2. Solution: we can assume the pyramid is built over the square −1 ≤x ≤1 and −1 ≤y ≤1. The cross section area at height h is A(h) = (2 −h)2. Therefore, V = Z 2 0 (2 −h)2 dh = 8/3 This is base area 4 times height 2 divided by 3. A solid of revolution is a surface enclosed by the surface obtained by rotating the graph of a function f(x) around the x axis. The area of the cross section at x of a solid of revolution is A(x) = πf(x)2. The volume of the solid is R b a πf(x)2 dx. 1 2 2 Find the volume of a round cone of height 2 and where the circular base has the radius 1. Solution. This is a solid of revolution obtained by rotation the graph of f(x) = x/2 around the x axes. The area of a cross section is πx2/4. Integrating this up from 0 to 2 gives Z 2 0 πx2/4 dx = x3 4 · 3|2 0 = 2π 3 . This is the height 2 times the base area π divided by 3. 3 Find the volume of a half sphere of radius 1. Solution: The area of the cross section at height h is π(1 −h2). 4 We rotate the graph of the function f(x) = sin(x) around the x axes. But now we cut out a slice of 60 = π/3 degrees out. Find the volume of the solid. Solution: The area of a slice without the missing piece is π sin2(x). The integral R π 0 sin2(x) dx is π/2 as derived in the lecture. Having cut out 1/6’th the area is (5/6)π sin2(x). The volume is R π 0 (5/6)π sin(x)2 dx = (5/6)π2/2. 3 Homework 1 Find the volume of the paraboloid for which the radius at position x is 1−x2 and x ranges from 0 to 1. 2 A catenoid is the surface obtained by rotating the graph of f(x) = cosh(x) around the x-axes. We have seen that the graph of f is the chain curve, the shape of a hanging chain. Find the volume of of the solid enclosed by the catenoid between x = −1 and x = 1. Hint. You might want to check first the identity 2 cosh(x)2 = 1 + cosh(2x) using the definition cosh(x) = (exp(x) + exp(−x))/2. 3 A tomato is given by z2 + x2 + 4y2 = 1. If we slice perpendicular to the y axes, we get a circular slice z2 + x2 ≤1 −4y2 of radius p 1 −4y2. a) Find the area of this slice. b) Determine the volume of the tomato. c) Fix yourself a tomato salad by cutting a fresh tomato into slices and eat it, except for one slice which you staple to your homework paper as proof that you really did it. 4 As we have seen in the movie of the first class, Archimedes was so proud of his formula for the vol-ume of a sphere that he wanted the formula on his tomb stone. He wrote the volume of a half sphere of radius 1 as the difference between the volume of a cylinder of radius 1 and height 1 and the volume of a cone of base ra-dius 1 and height 1. Relate the cross section area of the cylinder-cone complement with the cross section area of the sphere to recover his argument! If stuck, draw in the sand or soak in the bath tub for a while eating your tomato salad. There is no need to streak and scream ”Eureka” when the solution is found. 4 5 Volumes were among the first quantities, Mathematicians wanted to measure and compute. One problem on Moscow Eqyptian papyrus dating back to 1850 BC explains the general formula h(a2 + ab + b2)/3 for a truncated pyramid with base length a, roof length b and height h. a) Verify that if you slice the frustrum at height z, the area is (a + (b −a)z/h)2. b) Find the volume using calculus. Here is the translated formulation from the papyrus: 1 2 ”You are given a truncated pyramid of 6 for the vertical height by 4 on the base by 2 on the top. You are to square this 4 result 16. You are to double 4 result 8. You are to square 2, result 4. You are to add the 16, the 8 and the 4, result 28. You are to take one-third of 6 result 2. You are to take 28 twice, result 56. See it is 56. You will find it right”. 1Howard Eves, Great moments in mathematics, Volume 1, MAA, Dolciani Mathematical Expositions, 1980, page 10 2Image Source: papyri.html Math 1A: introduction to functions and calculus O. Knill and B. Lukoff, 2011 Lecture 22: Worksheet Volume Computation 1 Find the volume of the solid that is formed by rotating the graph of y = x2 around the x-axis, for 1 ≤x ≤3. 2 Find the volume of the solid that is formed by rotating the graph of y = x2 around the y-axis, for 1 ≤y ≤3. 3 Derive the formula for the volume of a sphere ( 4 3πr3). 4 Find the volume of the solid of revolution for which the radius at height z is 2 −|z| and −1 ≤z ≤1. 5 The solid of revolution for which the radius at position x is x4 + 1 and x ∈[−2, 2] is taken only above the xy plane as in the picture. Find the volume. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 23: Improper integrals In this lecture, we look at integrals on infinite intervals or integrals, where the function can get infinite at some point. These integrals are called improper integrals. The area under the curve can remain finite or become infinite. 1 What is the integral Z ∞ 1 1 x2 dx ? Since the anti-derivative is −1/x, we have −1 x |∞ 1 = −1/∞+ 1 = 1 . To justify this, compute the integral R b 1 1/x2dx = 1 −1/b and see that in the limit b →∞, the value 1 is achieved. In a previous lecture, we have seen a chocking example similar to the following one: 2 Z 1 −1 1 x2 dx = −1 x|1 −1 = −1 −1 = −2 . This does not make any sense because the function is positive so that the integral should be a positive area. The problem is this time not at the boundary −1, 1. The sore point is x = 0 over which we have carelessly integrated over. The next example illustrates the problem with the previous example better: 3 The computation Z 1 0 1 x2 dx = −1 x|1 0 = −1 + ∞. indicates that the integral does not exist. We can justify by looking at integrals Z 1 a 1 x2 dx = −1 x|1 a = −1 + 1 a which are fine for every a > 0. But this does not converge for a →0. Do we always have a problem if the function goes to infinity at some point? 4 Find the following integral Z 1 0 1 √x dx . 1 2 Solution: Since the point x = 0 is problematic, we integrate from a to 1 with positive a and then take the limit a →0. Since x−1/2 has the antiderivative x1/2/(1/2) = 2√x, we have Z 1 a 1 √x dx = 2√x|1 a = 2 √ 1 −2√a = 2(1 −√a) . There is no problem with taking the limit a →0. The answer is 2. Even so the region is infinite its area is finite. This is an interesting example. Imaging this to be a container for paint. We can fill the container with a finite amount of paint but the wall of the region has infinite length. 5 Evaluate the integral R 1 0 1/ √ 1 −x2 dx. Solution: The antiderivative is arcsin(x). In this case, it is not the point x = 0 which produces the difficulty. It is the point x = 1. Take a > 0 and evaluate Z 1−a 0 1 √ 1 −x2 dx = arcsin(x)|1−a 0 = arcsin(1 −a) −arcsin(0) . Now arcsin(1 −a) has no problem at limit a →0. Since arcsin(1) = π/2 exists. We get therefore the answer arcsin(1) = π/2. 6 Rotate the graph of f(x) = 1/x around the x-axes and compute the volume of the solid between 1 and ∞. The cross section area is π/x2. If we look at the integral from 1 to a fixed R, we get Z R 1 π x2 dx = −π x|R 1 = −π/R + π . This converges for R →∞. The volume is π. This famous solid is called Gabriels trumpet. This solid is so prominent because if you look at the surface area of the small slice, then it is larger than dx2π/x. The total surface area of the trumpet from 1 to R is therefore larger than R R 1 2π/x dx = 2π(log(R) −log(1)). which goes to infinity. We can fill the trumpet with a finite amount of paint but we can not paint its surface. 3 Finally, lets look at the following example 7 Evaluate the integral R ∞ 0 sin(x) dx. Solution. There is no problem at the boundary 0 nor at any other point. We have to investigate however, what happens at ∞. Therefore, we look at the integral R b 0 sin(x) dx = −cos(x)|b 0 = 1 −cos(b). We see that the limit b →∞ does not exist. The integral fluctuates between 0 and 2. The next example leads to a topic in a follow-up course. It is not covered here, but could make you curious: 8 What about the integral I = Z ∞ 0 sin(x) x dx ? Solution. The anti derivative is the Sine integral Si(x) so that we can write R b 0 sin(x)/x dx = Si(b). It turns out that the limit b →∞exists and is equal to π/2 but this is a topic for a 4 second semester course like Math 1b. The integral can be written as an alternating series, which converges and there are many ways to compute it: 1 Lets summarize the two cases of improper integrals: infinitely long intervals and a point where the function becomes infinite. 1) To investigate the improper integral R ∞ a f(x) dx we look at the limit R b a f(x) dx for b →∞. 1) To investigate improper integral R b 0 f(x) dx where f(x) is not continuous at 0, we take the limit R b a f(x) dx for a →0. Homework 1 Evaluate the integral R 2 1 5 √x−1 + cos(πx) dx. 2 Evaluate the following integrals a) R 1 0 x/ √ 1 −x2 dx. b) R 1 0 1/ √ 1 −x2 dx. Hint: For a) think about the chain rule d/dxf(g(x)) = f ′(g(x))g′(x) 3 Evaluate the integral R 4 −3(x2)1/3 dx. To make sure that the integral is fine, check whether R 0 −3 and R 4 0 work. 4 The integral R 1 −2 1/x dx does not exist. We can however take a positive b > 0 and look at Z −b −2 1/x dx + Z 1 b 1/x dx = log |b| −log | −2|) + (log |1| −log |b|) = log(2) . This value is called the Cauchy principal value of the integral. Find the principal value of Z 5 −4 3/x3 dx using the same process as before, by cutting out [−a, a] and then taking the limit a →0. 5 Could we have given a principal value integral value to R 1 −1 1 x2 dx? If yes, find the value. If not, tell why not. 1Hardy, Mathematical Gazette, 5, 98-103, 1909. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 23: Improper integrals In this lecture, we look at integrals on infinite intervals or integrals, where the function can get infinite at some point. These integrals are called improper integrals. The area under the curve can remain finite or become infinite. 1 What is the integral Z ∞ 1 1 x2 dx ? Since the anti-derivative is −1/x, we have −1 x |∞ 1 = −1/∞+ 1 = 1 . To justify this, compute the integral R b 1 1/x2dx = 1 −1/b and see that in the limit b →∞, the value 1 is achieved. In a previous lecture, we have seen a chocking example similar to the following one: 2 Z 1 −1 1 x2 dx = −1 x|1 −1 = −1 −1 = −2 . This does not make any sense because the function is positive so that the integral should be a positive area. The problem is this time not at the boundary −1, 1. The sore point is x = 0 over which we have carelessly integrated over. The next example illustrates the problem with the previous example better: 3 The computation Z 1 0 1 x2 dx = −1 x|1 0 = −1 + ∞. indicates that the integral does not exist. We can justify by looking at integrals Z 1 a 1 x2 dx = −1 x|1 a = −1 + 1 a which are fine for every a > 0. But this does not converge for a →0. Do we always have a problem if the function goes to infinity at some point? 4 Find the following integral Z 1 0 1 √x dx . 1 2 Solution: Since the point x = 0 is problematic, we integrate from a to 1 with positive a and then take the limit a →0. Since x−1/2 has the antiderivative x1/2/(1/2) = 2√x, we have Z 1 a 1 √x dx = 2√x|1 a = 2 √ 1 −2√a = 2(1 −√a) . There is no problem with taking the limit a →0. The answer is 2. Even so the region is infinite its area is finite. This is an interesting example. Imaging this to be a container for paint. We can fill the container with a finite amount of paint but the wall of the region has infinite length. 5 Evaluate the integral R 1 0 1/ √ 1 −x2 dx. Solution: The antiderivative is arcsin(x). In this case, it is not the point x = 0 which produces the difficulty. It is the point x = 1. Take a > 0 and evaluate Z 1−a 0 1 √ 1 −x2 dx = arcsin(x)|1−a 0 = arcsin(1 −a) −arcsin(0) . Now arcsin(1 −a) has no problem at limit a →0. Since arcsin(1) = π/2 exists. We get therefore the answer arcsin(1) = π/2. 6 Rotate the graph of f(x) = 1/x around the x-axes and compute the volume of the solid between 1 and ∞. The cross section area is π/x2. If we look at the integral from 1 to a fixed R, we get Z R 1 π x2 dx = −π x|R 1 = −π/R + π . This converges for R →∞. The volume is π. This famous solid is called Gabriels trumpet. This solid is so prominent because if you look at the surface area of the small slice, then it is larger than dx2π/x. The total surface area of the trumpet from 1 to R is therefore larger than R R 1 2π/x dx = 2π(log(R) −log(1)). which goes to infinity. We can fill the trumpet with a finite amount of paint but we can not paint its surface. 3 Finally, lets look at the following example 7 Evaluate the integral R ∞ 0 sin(x) dx. Solution. There is no problem at the boundary 0 nor at any other point. We have to investigate however, what happens at ∞. Therefore, we look at the integral R b 0 sin(x) dx = −cos(x)|b 0 = 1 −cos(b). We see that the limit b →∞ does not exist. The integral fluctuates between 0 and 2. The next example leads to a topic in a follow-up course. It is not covered here, but could make you curious: 8 What about the integral I = Z ∞ 0 sin(x) x dx ? Solution. The anti derivative is the Sine integral Si(x) so that we can write R b 0 sin(x)/x dx = Si(b). It turns out that the limit b →∞exists and is equal to π/2 but this is a topic for a 4 second semester course like Math 1b. The integral can be written as an alternating series, which converges and there are many ways to compute it: 1 Lets summarize the two cases of improper integrals: infinitely long intervals and a point where the function becomes infinite. 1) To investigate the improper integral R ∞ a f(x) dx we look at the limit R b a f(x) dx for b →∞. 1) To investigate improper integral R b 0 f(x) dx where f(x) is not continuous at 0, we take the limit R b a f(x) dx for a →0. Homework 1 Evaluate the integral R 2 1 5 √x−1 + cos(πx) dx. 2 Evaluate the following integrals a) R 1 0 x/ √ 1 −x2 dx. b) R 1 0 1/ √ 1 −x2 dx. Hint: For a) think about the chain rule d/dxf(g(x)) = f ′(g(x))g′(x) 3 Evaluate the integral R 4 −3(x2)1/3 dx. To make sure that the integral is fine, check whether R 0 −3 and R 4 0 work. 4 The integral R 1 −2 1/x dx does not exist. We can however take a positive b > 0 and look at Z b −2 1/x dx + Z 1 a 1/x dx = log |b| −log | −2|) + (log |1| −log |b|) = log(2) . This value is called the Cauchy principal value of the integral. Find the principal value of Z 5 −4 3/x3 dx using the same process as before, by cutting out [−a, a] and then taking the limit a →0. 5 Could we have given a principal value integral value to R 1 −1 1 x2 dx? If yes, find the value. If not, tell why not. 1Hardy, Mathematical Gazette, 5, 98-103, 1909. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 23: Worksheet Inproper integrals 1 Find the value of the improper integral Z ∞ 1 1 x11 dx 2 Find the following improper integral Z 1 0 1 √1 −x . 3 We have met the Maria Agnesi function f(x) = 1 1 + x2 early in the course already. Evaluate the integral I = Z ∞ −∞ 1 1 + x2 dx The function g(x) = 1 I 1 1+x2 is a probability distribution called Cauchy distribution. It is a nonzero function which has the property that R ∞ −∞g(x) dx = 1. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 24: Applications of integration You have seen these integration applications: • the computation of area • the computation of volume • position from acceleration • cost from marginal cost Here are some more: • probabilities and distributions • averages and expectations • finding moments of inertia • work from power Probability In probability theory functions are used as observables or to define probabilities. Assuming our probability space to be the real line, an interval [a, b] is called an event. Given a nonnegative function f(x) which has the property that R ∞ −∞f(x) dx = 1, we call P[A] = Z b 0 f(x) dx the probability of the event. The function f(x) is called the probability density function. The most famous and most important probability density is the normal distribution: The normal distribution has the density f(x) = 1 √ 2πe−x2/x . It is the distribution which appears most often if data can take both positive and negative values. The reason why it appears so often is that if one observes different unrelated quantities with the same statistical properties, then their sum, suitably normalized becomes the normal distribution. If we measure errors for example, then these errors often have a normal distribution. a b 1 The probability density function of the exponential distribution is defined as f(x) = e−x for x ≥0 and f(x) = 0 for x < 0. It is used to used measure lengths of arrival times like the time until you get the next phone call. The density is zero for negative x because there is no way we can travel back in time. What is the probability that you get a phone call between times x = 1 and times x = 2 from now? The answer is R 2 1 f(x) dx. 1 2 a b Assume f is a probability density function (PDF). The antiderivative F(x) = R x −∞f(t) dt is called the cumulative distribution function (CDF). 2 For the exponential function the cumulative distribution function is Z x −∞ f(x) dx = Z x 0 f(x) dx = −e−x|x 0 = 1 −e−x . The probability density function f(x) = 1 π 1 1+x2 is called the Cauchy distribution. 3 Find its cumulative distribution function. Solution: F(x) = Z x −∞ f(t) dt = 1 π arctan(x)|x −∞= ( 1 π arctan(x) + 1 2) . a b Average Here is an example for computing the average. 4 Assume the level in a honey jar over [0, 2π] containing crystallized honey is given by a function f(x) = 3 + sin(3x)/5 + x(2π −x)/10. In order to restore the honey, it is placed into hot water. The honey melts to its normal state. What height does it have? Solution: The average height is R 2π 0 f(x) dx/(2π) which is the area divided by the base length. In probability theory we would call f(x) a random variable and the average of f with E[f] the expectation. 3 Moment of inertia If we spin a wire of radius L of mass density f(x) around an axes, the moment of inertia is defined as I = R L 0 x2f(x) dx. The significance is that if we spin it with angular velocity w, then the energy is Iw/2. 5 Assume a wire has density 1 + x and length 3. Find its moment of inertia. Solution: 6 Flywheels have a comeback for powerplants to absorb energy. If there is not enough power, the flywheels are charged, in peak times, the energy is recovered. They work with 80 percent efficiency. Assume a flywheel is a cylinder of radius 1, density 1 and height 1, then the moment of inertia integral is R 1 0 z2f(z) dx, where f(z) is the mass in distance z. Work from power If P(t) is the amount of power produced at time t, then R T 0 P(t) dt is the work=energy produced in the time interval [0, T]. Energy is the anti-derivative of power. 7 Assume a power plant produces power P(t) = 1000 + exp(−t) + t2 −t. What is the energy produced from t = 1 to t = 10? Solution. 4 Wouldn’t be nice to have one of those bikes with interactive training environments in the gym, allowing to ride in the Peruvian or Swiss Mountains, the California coast or in the Italian Tuscany? Additionally, there should be some computer game features, racing other riders through beaches, deserts or Texan highways (could be on google earth). Training would be so much more enter-taining. Business opportunities everywhere. The first offering such training equipment will make a fortune. Until then we are stuck with TV programs which really suck. Homework 1 The probability distribution which describes the time you have to wait for your next email is f(x) = 3e−3x. What is the probability that you get your next email in the next 2 hours, that is between x = 0 and x = 2? 2 Assume the probability distribution for the waiting time to the next warm day is f(x) = (1/4)e−x/4, where x has days as unit. What is the probability to get a warm day between tomorrow and after tomorrow that is between x = 1 and x = 2? 3 A rod modeled over the interval [0, 4] has temperature f(x) = 5 + x2 −3x at position x. Find the average temperature. 4 A CD Rom has radius 6. If we would place the material at radius x onto one point, we get a density of f(x) = 2πx. Find the moment of inertia I of the disc. If we spin it with an angular velocity of w = 20 rounds per second. Find the energy E = Iw2/2. Without credit: Explode a CD: Careful! 5 a) You are on a stationary bike in the Hemenway gym and pedal with power P(t) = 200 + 100 sin(10πt) − t 300 + t2 19440 (in Watts=W). The periodic fluctuations come from a hilly route. The linear term is the ”tiring effect” and the quadratic term is due to endorphins kicking in eventually. What energy (Joules J=W s) have you produced in the time t ∈[0, 1800] (s=seconds)? b) Since we do math not physics, we usually ignore all the units but this one is just too much fun. If you divide the result by 4184, you get kilo calories = food calories. Eating an apple gives you about 80 food calories. How many apples can you eat after your half hour workout, just to get even? Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 24: Worksheet Applications of integration 1 Find the cumulative distribution function F(x) = Z x −∞f(t) dt . of the exponential distribution in the case f(x) = 2 exp(−2x). 2 Find the moment of inertia of a rod which has density f(x) = x and length 10. Z L 0 x2f(x) dx . 3 A light bulb produces 100W. How much energy in kw/Hours does it use in 1 year? Assume you pay 10 cents for each kW/h. How much does it cost? Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 25: Related rates Before we continue with integration, we include a short flash-back on differentiation. This allows us to solidify the chain rule d dxf(g(x)) = f ′(g(x))g′(x) which will be very useful for the integration technique called ”substitution”. Since the chain rule is often perceived as a difficult concept in calculus, it is good to come back to it again. We take the opportunity also to review a bit our differentiation skills and to take some fresh breath before launching into more advanced integration techniques. 1 Assume we inflate a balloon and pump 5 volume units per unit time into it. If the balloon has radius 7, what is the rate of change of the radius? Solution. Let V (r) be the volume and r(t) the radius at time t. Since V (r(t)) = 4πr(t)3/3 , we have by the chain rule 5 = d/dtV (r(t)) = 4πr(t)2r′(t) . This relation allows us to compute r′(t) = 5/(4πr2) = 5/(4π72). 2 Hydrophilic water gel spheres made from polyacrylamide polymer can expand 300 times their original size as you see in class. Assume they have initially a diameter of 1 (cm) and that they expand in 10 hours to its 300 fold volume. Find the rate of change of the radius in time when they have a volume of 100 (cm3). Solution. We have the same rule V = 4πr3/3 . The problem gives us d/dtV (r(t)) = 300/10 = 30. The rest is now the same as in the previous problem: 30 = 4πr2r′. Since r = 100 we get r′ = 30/(4π1002). 3 The upper part of a wine glass has a shape y = x2 with 0 ≤y ≤2. We assume the glass is half full, meaning that the wine level is at y = 1. We taste the wine with 1 ml/sec using a straw, ignoring any political and behavioral correctness. How fast does the wine level sink at that moment? Solution: The area of the wine layer at height y is A(y) = x2π = yπ. The volume is V (y) = R y 0 yπ dy = y2π/2 . We know −1 = d/dtV (y(t)) = V ′(y)y′(t) = πyy′(t) so that y′(t) = −1/(πy) and for y = 1 this is −1/π. 1 2 4 A person of height 6 feet is located at x = 6 and walks with constant speed 1. A lamp at x = 0 is at height 10 feet. With what speed does the shadow of the person proceed on the floor? Solution: If the person is at position x, the shadow’s length L satisfies L/6 = (L + x)/10 which is L = 9. The relation L/6 = (L + x)/10 means L = 3x/2 so that L′ = 3x′/2 = 3/2. 5 Romeo and Juliet have meet secretly at position (0, 0) and rush home. Romeo runs with speed 4 meters/seconds to the east. Assume their distance satisfies l(t) = t3. After 10 seconds, they wave back to each other. With what speed does Juliet run at this time? Solution. What do we know? x(t) = 4t is the position of Romeo and l(t) = t3. If y(t) is the y position of Juliet, the law we use is Pythagoras l2 = x2 + y2 so that y(t) = √ l2 −x2 and y(10) = √1000 −100 = √ 900 = 30. Now differentiate the law to get 2ll′ = 2xx′+2yy′. We know all quantities at time t = 10: we know l = 1000, l′ = 300, x = 40, x′ = 4, y = 30 and compute y′ = (2000 ∗300 −80 ∗4)/60 = 29984/3. 3 We have seen the ladder example twice already: 6 A ladder has length 1. Assume slips on the ground away with constant speed 2 in the x-direction. What is the speed of the top part of the ladder sliding down the wall at the time when x = y? Solution We know x′(t) = 2 and that x(t), y(t) are related by x2(t) + y2(t) = 1 . Differentiation gives 2x(t)x′(t) + 2y(t)y′(t) = 0. We get y′(t) = −x′(t)x(t)/y(t) = 2 · 1 = 1. 7 A kid slides down a slide of the shape y = 2/x . Assume at height y = 2 we have dy/dt = −7. What is dx/dt? Solution: differentiate the relation to get y′ = −2x′/x2. At y = 2 we have x = 1. Now solve for x′ to get x′ = −y′x2/2 = 7/2. Image source: 8 A canister of oil releases oil at a constant rate 5. With what rate does the radius of the oil spill increase, when the radius is 1? Solution. We have A(r) = r2π and so 5 = A′(r) = 2rr′π. Solving for r′ gives r′ = 5/(2rπ) which is 5/(2π). Related rates problems link quantities by a rule . These quantities can depend on time. To solve a related rates problem, differentiate the rule with respect to time and solve for the unknown quantity. Related rates problems are not so easy. The difficulty comes from the fact that they are often ”word problems” which first have to be parsed. We have to find the rule and differentiate it. In all the problems on this handout, the rule is boxed. It is important to understand which variables depend on time. If a term x3 appears for example and x depends on time, then d/dtx3 = 3x2x′. 4 Homework 1 The ideal gas law pV = T relates pressure p and vol-ume V and temperature T. Assume the temperature T = 50 is fixed and the volume is at V = 2 and de-creased by V ′ = −3. Find the rate p′ with which the pressure increases. 2 Assume the total production rate P of a new tablet computer product for kids is constant 100 and given by the famous Cobb-Douglas formula P = L1/3K2/3 where L = 64 is the labor and K = 125 is the cost. Assume labor is increased at a rate L′ = 2. What is the cost change K′? 3 You observe an airplane at height h = 10′000 meters directly above you and see that it moves with rate φ′ = 5 degree per second (which is 5π/180 radiants per second). What is the speed x′ of the airplane directly above you where x = 0? Hint: Use tan(φ) = x/h and make a picture to figure out what φ is. 4 An isosceles triangle with base 2a and height h has fixed area A = ah = 1 . Assume the height is decreased by a rate h′ = −2. With what rate does a increase if h = 1/2? a h 5 There are cosmological models which see our universe as a four dimensional sphere which expands in space time. Assume the volume V = π2r4/2 increases at a rate d/dtV (r(t)) = 100π2r2. What is r′ if the current radius is r = 47 (billion light years). Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 25: Worksheet Related rates 1 An underwater oil spill releases oil at the constant amount. The area A(r) of the oil increases with A′(r(t)) = 2. If the radius is r = 4, what is the rate of change of r? 2 The resistance R, voltage U and current I are related by U = RI . Assume the temperature increases, the resistance R(t) increases by a constant amount R′ = 2. If the voltage stays constant U = 4 what is the rate of change of I? V R I I Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 26: Implicit differentiation We have seen an implicit differentiation example in the Valentines day lecture and will repeat this topic more. Implicit differentiation is also crucial to find the derivative of inverse functions. We will review this here because this will give us handy tools for integration. The chain rule, related rates and implicit differentiation are all the same concept, but viewed from different angles. You can see implicit differentiation as a special case of related rates where one of the quantities is ”time” meaning that this is the variable with respect to which we differentiate. 1 Points (x, y) in the plane which satisfy x2 + 9y2 = 10 form an ellipse. Find the slope of the tangent line at the point (1, 1). Solution: We want to know the derivative dy/dx. We have 2x + 18yy′ = 0. Using x = 1, y = 1 we see y′ = −2x/(18y) = −1/9. Remark. We could have looked at this as a related rates problem where x(t), y(t) are related and x′ = 1 Now 2xx′ + 9 · 2yy′ = 0 allows to solve for y′ = −2xx′/(9y) = −2/9. 2 The points (x, y) which satisfy the equation x4 −3x2 + y2 = 0 forms a figure 8 called lemniscate of Gerono. It contains the point (1, √ 2). Find the slope of the curve at that point. Solution: We differentiate the law describing the curve with respect to x. This gives 5x3 −6x + 2yy′ = 0 We can now solve for y′ = (6x −5x3)/(2y) = 1/2. 1 2 3 The Valentine equation (x2 + y2 −1)3 −x2y3 = 0 contains the point (1, 1). Near (1, 1), we have y = y(x) so that (x2 + y(x)2 −1)3 −x2y(x)3 = 0. Find y′ at the point x = 1. Solution Take the derivative 0 = 3(x2 + y2 −1)(2x + 2yy′) −2xy3 −x23y2y′(x) and solve for y′ = −3(x2 + y2 −1)2x −2xy3 3(x2 + y2 −1)2y −3x2y2 . For x = 1, y = 1, we get −4/3. 4 The energy of a pendulum with angle x and angular velocity y is y2 −cos(x) = 1 is constant. What is y′? We could solve for y and then differentiate. Simpler is to differentiate directly and get yy′ + sin(x) = 0 so that y′ = −sin(x)/y. At the point (π/2, 1) for example we have y′ = −1. 3 What is the difference between related rates and implicit differentiation? Implicit differentiation is the special case of related rates where one of the variables is time. Derivatives of inverse functions Implicit differentiation has an important application: it allows to compute the derivatives of inverse functions. It is good that we review this, because we can use these derivatives to find anti-derivatives. 5 Find the derivative of log(x) by differentiating exp(log(x)) = x. Solution: 1 = d dxx = d dx exp(log(x)) = exp(log(x)) d dx log(x) = x log′(x) . Solve for log′(x) = 1/x. Since the derivative of log(x) is 1/x. The anti-derivative of 1/x is log(x) + C. 6 Find the derivative of arccos(x) by differentiating cos(arccos(x)) = x. Solution: 1 = d dxx = d dx cos(arccos(x)) = −sin(arccos(x)) arccos′(x) = − p 1 −cos2(arccos(x)) arccos′(x) = − √ 1 −x2 arccos′(x) . Solving for arccos′(x) = −1/ √ 1 −x2. The anti-derivative of arccos(x) is −1/ √ 1 −x2. 7 Find the derivative of arctan(x) by differentiating tan(arctan(x)) = x. Solution: This is a derivative which we have seen several times by now. We use the identity 1/ cos2(x) = tan2(x) + 1 to get 1 = d dxx = d dx tan(arctan(x)) = 1 cos2(x) arctan′(x) = (1 + tan2(arctan(x))) arctan′(x) . 4 Solve for arctan′(x) = 1/(1 + x2). The anti-derivative of arctan(x) is 1/(1 + x2). 8 Find the derivative of f(x) = √x by differentiating (√x)2 = x. Solution: 1 = d dxx = d x(√x)2 = 2√xf ′(x) so that f ′(x) = 1/(2√x). Homework 1 The equation y2 = x2−x defines the graph of the function f(x) = √ x2 −x. Find the slope of the graph at x = 2 directly by differentiating f. Then use the implicit differentiation method and differentiate y2 = x2 −x assuming y(x) is a function of x and solving for y′. 2 The equation x2 + y2 = 5 defines a circle. Find the slope of the tangent at (1, 2). 3 The equation x100 + y100 = 1 + 2100 defines a curve which looks close to a square. Find the slope of the curve at (2, 1). 4 Derive again the derivative of arccot(x) as we did before in this course and also during the first midterm. 5 a) The relation sin(x −y) −2 cos((π/2)xy) = 0 relates x(t) and y(t). Assume x′ = 1 at (1, 1) what is y′? This is a related rates problem. b) Now do it directly. Since x′ = 1 we can use x as the variable. Find y′(x) by implicit differentiation. You should get the same result as in a). Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 26: April First Worksheet Implicit differentiation 1 Find the slope of y′(x) if 2x3 −y3 = y at the point (1, 1). 2 Find the derivative of y(x) = x1/5 by differentiating y5 = x. 3 The equation y = x relates two quantities and defines y in terms of x. Assume x = 2 find d dxy . Hint. This problem tries hard to be a April first joke but does not quite succeed. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 27: Review for second midterm Major points The intermediate value theorem assures that there is x ∈(a, b) with f ′(x) = (f(b) −f(a))/(b −a). A special case is Rolle’s theorem, where f(b) = f(a). Catastrophes are parameter values where a local minimum disappears. Typically the system jumps then to a lower minimum. Definite integrals F(x) = R x 0 f(x) dx are defined as a limit of Riemann sums Sn/n. A function F(x) satisfying F ′ = f is called the anti-derivative of f. The general anti-derivative is F + C where C is a constant. The fundamental theorem of calculus tells d/dx R x 0 f(x) dx = f(x) and R x 0 f ′(x) dx = f(x) −f(0). The integral R b a g(x) −f(x) dx is the signed area between the graphs of f and g. Places, where f < g are counted negative. The integral R b a A(x) dx is a volume if A(x) is the area of a slice of the solid perpendicular to a point x on an axes. Write improper integrals as limits of definite integrals R ∞ 1 f(x) dx = limR→∞ R R 1 f(x) dx. We similarly treat points, where f is discontinuous. Besides area, volume, total cost, or position, we can compute averages, inertia or work using integrals. If x, y are related by F(x(t), y(t)) = 0 and x(t) is known we can compute y′(t) using the chain rule. This is related rates. If f(g(t)) is known we can compute g′(x) using the chain rule. This works for inverse functions. This is implicit differentiation. To determine the catastrophes for a family fc(x) of functions, determine the critical points in depence of c and find values c, where a critical point changes from a local minimum to a local maximum. 1 2 Important integrals cos(x) sin(x). sin(x) −cos(x). tan(x) 1/ cos2(x). 1/(1 + x2) arctan(x). 1/ √ 1 −x2 arcsin(x) exp(x) exp(x) log(x) x log(x) −x 1/x log(x) −1/(1 + x2) arccot(x). −1/ √ 1 −x2 arccos(x) Improper integrals R ∞ 1 1/x2 dx Prototype of first type improper integral which exists. R ∞ 1 1/x dx Prototype of first type improper integral which does not exist. R 1 0 1/x dx Prototype of second type improper integral which does not exist. R 1 0 1/√x dx Prototype of second type improper integral which does exist. The fundamental theorem d dx R x 0 f(x) dx = f(x) R x 0 f ′(x) dx = f(x) −f(0). This implies R b a f ′(x) dx = f(b) −f(a) Without limits of integration, we call R f(x) dx the anti derivative. It is defined up to a constant. For example R sin(x) dx = −cos(x) + C. Applications Calculus applies directly if there are situations where one quantity is the derivative of the other. function anti derivative acceleration velocity velocity position function area under the graph length of cross section area of region area of cross section volume of solid marginal prize total prize power work probability density function cumulative distribution function Tricks Whenever dealing with an area or volume computation, make a picture. In related rates problems, make sure you understand what are variables and what are constants. For volume computations, find the area of the cross section A(x) and integrate. For area computations find the length of the slice f(x) and integrate. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 27: Review Problems Definite integral 1 The following integral defines the area of a region. Draw it: Z π π/2 x −sin(x) dx . Catastrophes 2 Lets look at the family of functions fc(x) = x5 + cx3. You see three graphs. They display the function for c = −1, c = 0 and c = 1. What can you say about catastrophes? -1.0 -0.5 0.5 1.0 -0.15 -0.10 -0.05 0.05 0.10 0.15 -1.0 -0.5 0.5 1.0 -0.4 -0.2 0.2 0.4 -1.0 -0.5 0.5 1.0 -10 -5 5 10 Area 3 Find the area of the region bound by y = 2 −x, x = y, y = 0 and y = 1. Volumes 4 If we rotate the witch of Agnesi y = (1 + x2)−1 around the x axes, we obtain a solid. Find its volume. Hint. To find the integral, compute the derivative of x/(1 + x2) and get inspired. Related Rates 5 The curve x2 −y2 = 3y is and example of a hyperbola. If x(t) = 2 + t. Find the related rate y′ near (2, 1). -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6 Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 28: Substitution If we differentiate the function sin(x2) and use the chain rule, we get cos(x2)2x. By the fundamental theorem of calculus, the anti derivative of cos(x2)2x is sin(x2). We know therefore Z cos(x2)2x dx = sin(x2) + C . Spotting the chain rule How can we see the integral without knowing the result already? Here is a very important case: If we can spot that f(x) = g(u(x))u′(x), then the anti derivative of f is G(u(x)) where G is the anti derivative of g. 1 Find the anti derivative of ex4+x2(4x3 + 2x) . Solution: The derivative of the inner function is to the right. 2 Find Z √ x5 + 1x4 dx . Solution. The derivative of x5 + 1 is 5x4. This is almost what we have there but the constant can be adapted. The answer is (1/5)(x5 + 1)3/2. 3 Find the anti derivative of log(x) x . Solution: The derivative of log(x) is 1/x. The antiderivative is log(x)2/2. 4 Find the anti derivative of cos(sin(x2)) cos(x)2x . Solution. We see the derivative of sin(x2) appear on the right. Therefore, we have sin(sin(x2)). In the next three examples, substitution is actually not necessary. You can just write down the anti derivative, and adjust the constant. It uses the following ”speedy rule”: If R f(ax + b) dx = F(ax + b)/a where F is the anti derivative of f. 1 2 5 R √x + 1 dx. Solution: (x + 1)3/2(2/3). 6 R 1 1+(5x+2)2 dx. Solution: arctan(5x + 2)(1/5). Doing substitution Spotting things is sometimes not easy. The method of substitution helps to formalize this. To do so, identify a part of the formula to integrate and call it u then replace an occurrence of u′dx with du. Z f( u(x) ) u’(x) dx = Z f( u ) du . Here is a more detailed description: replace a prominent part of the function with a new variable u, then use du = u′(x)dx to replace dx with du/u′. We aim to end up with an integral R g(u) du which does not involve x anymore. Finally, after integration of this integral, replace the variable u again with the function u(x). The last step is called back-substitution. 7 Find the anti-derivative Z log(log(x))/x dx . Solution Replace log(x) with u and replace u′dx = 1/xdx with du. This gives R log(u) du = u log(u) −u = log(x) log log(x)) −log(x). 8 Solve the integral Z x/(1 + x4) dx . Solution Substitute u = x2, du = 2xdx to get (1/2) R du/(1 + u2) du = (1/2) arctan(u) = (1/2) arctan(x2). 9 Solve the integral Z sin(√x)/√x . Here are some examples which are not so straightforward: 10 Solve the integral Z sin3(x) dx . 3 Solution. We replace sin2(x) with 1 −cos2(x) to get Z sin3(x) dx = Z sin(x)(1 −cos2(x)) dx = −cos(x) + cos3(x)/3 . 11 Solve the integral Z x2 + 1 √x + 1 dx . Solution: Substitute u = √x + 1. This gives x = u2 −1, dx = 2udu and we get R 2(u2 − 1)2 + 1 du. 12 Solve the integral Z x3 √ x2 + 1 dx . Trying u = √ x2 + 1 but this does not work. Try u = x2 + 1, then du = 2xdx and dx = du/(2√u −1). Substitute this in to get Z √u −1 3 2√u −1√u du = Z (u −1) 2√u = Z u1/2/2−u−1/2/2 du = u3/2/3−u1/2 = (x2 + 1)3/2 3 −(x2+1)1/2 . Definite integrals When doing definite integrals, we could find the antiderivative as described and then fill in the boundary points. Substituting the boundaries directly accelerates the process since we do not have to substitute back to the original variables: Z b a g(u(x))u′(x) dx = Z u(b) u(a) g(u) du . Proof. This identity follows from the fact that the right hand side is G(u(b)) −G(u(a)) by the fundamental theorem of calculus. The integrand on the left has the anti derivative G(u(x)). Again by the fundamental theorem of calculus the integral leads to G(u(b)) −G(u(a)). Top: To keep track which bounds we consider it can help to write R x=b x=a f(x) dx. 13 Find the anti derivative of R 2 0 sin(x3 −1)x2 dx. Solution. Z x=2 x=0 sin(x3 + 1)x2 dx . Solution: Use u = x3 + 1 and get du = 3x2dx. We get Z u=7 u=1 sin(u)du/3 = (1/3) cos(u)|7 1 = [−cos(7) + cos(1)]/3 . Also here, we can see the integrals directly To integrate f(Ax + B) from a to b we get [F(Ab + B) −F(Aa + B)]/A, where F is the anti-derivative of f. 14 R 1 0 1 5x+1 dx = [log(u)]/5|6 1 = log(6)/5. 15 R 5 3 exp(4x −10) dx = [exp(10) −exp(2)]/4. 4 Homework 1 Find the following anti derivatives. a) R 5x sin(x2) dx b) R ex5+x(x4 + 1/5) dx c) cos(cos(x)) sin(x) d) etan(x)/ cos2(x). 2 Find the following definite integrals. a) R 5 3 √ x5 + x(x4 + 1/5) dx b) R √π 0 sin(x2)x dx. c) R e 1/e √ log(x) x dx. d) R 1 0 x/ √ 1 + x2 dx. 3 a) Find the integral R 1 0 3x √ 1 −x4 dx using a substitution and interpreting the result using a known area. b) Find the moment of inertia of a rod with density f(x) = √ x3 + 1 between x = 0 and x = 4. 4 a) Integrate Z 1 0 arcsin(x) √ 1 −x2 dx . b) Find the definite integral Z 6e e dx p log(x)x . 5 a) Find the indefinite integral Z x5 √ x2 + 1 dx . b) Find the anti-derivative of f(x) = 1 x(1 + log(x)2) . Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 28: Worksheet Substitution 1 Z sin(2x + 3) dx 2 Z 1 (x + 8)5 dx 3 Z log(5x) x dx 4 Z x √ x2 + 1 dx 5 Z ex (ex + 5)2 ; dx Here is an situation, where substitution appears in an application lets look at the probability density function. The integral m = Z ∞ −∞xf(x) dx is called the mean of the distribution. 6 Find the mean of the probability density function f(x) = 1 √πe−(x−3)2/2 . Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 29: Integration by parts If we integrate the product rule (uv)′ = u′v +uv′ we obtain an integration rule called integration by parts. It is a powerful tool, which complements substitution. As a rule of thumb, always try first to simplify a function and integrate directly, then give substitution a first shot before trying integration by parts. R u(x) v’ (x)dx = u(x)v(x) − R u′(x)v(x) dx. 1 Find R x sin(x) dx. Solution. Lets identify the part which we want to differentiate and call it u and the part to integrate and call it v′. The integration by parts method now proceeds by writing down uv and subtracting a new integral which integrates u′v : Z x sin(x) dx = x (−cos(x)) − Z 1 (−cos(x) ) dx = −x cos(x) + sin(x) + C dx . 2 Find R xex dx. Solution. Z x exp(x) dx = x exp(x) − Z 1 exp(x) dx = x exp(x) −exp(x) + C dx . 3 Find R log(x) dx. Solution. There is only one function here, but we can look at it as log(x) · 1 Z log(x) 1 dx = log(x)x − Z 1 xx dx = x log(x) −x + C . 4 Find R x log(x) dx. Solution. Since we know from the previous problem how to integrate log we could proceed like this. We would get through but what if we do not know? Lets differentiate log(x) and integrate x: Z log(x) x dx = log(x)x2 2 − Z 1 x x2 2 dx which is log(x)x2/2 −x2/4. We see that it is better to differentiate log first. 5 Marry go round: Find I = R sin(x) exp(x) dx. Solution. Lets integrate exp(x) and differentiate sin(x). = sin(x) exp(x) − Z cos(x) exp(x) dx . Lets do it again: = sin(x) exp(x) −cos(x) exp(x) − Z sin(x) exp(x) dx. We moved in circles and are stuck! Are we really. We have derived an identity I = sin(x) exp(x) −cos(x) exp(x) −I which we can solve for I and get I = [sin(x) exp(x) −cos(x) exp(x)]/2 . 1 2 Tic-Tac-Toe Integration by parts can bog you down if you do it sev-eral times. Keeping the order of the signs can be daunt-ing. This is why a tabular integration by parts method is so powerful. It has been called ”Tic-Tac-Toe” in the movie Stand and deliver. Lets call it Tic-Tac-Toe therefore. 6 Find the anti-derivative of x2 sin(x). Solution: x2 sin(x) 2x −cos(x) ⊕ 2 −sin(x) ⊖ 0 cos(x) ⊕ The antiderivative is −x2 cos(x) + 2x sin(x) + 2 cos(x) + C . 7 Find the anti-derivative of (x −1)3e2x. Solution: (x −1)3 exp(2x) 3(x −1)2 exp(2x)/2 ⊕ 6(x −1) exp(2x)/4 ⊖ 6 exp(2x)/8 ⊕ 0 exp(2x)/16 ⊖ The anti-derivative is (x −1)3e2x/2 −3(x −1)2e2x/4 + 6(x −1)e2x/8 −6e2x/16 + C . 8 Find the anti-derivative of x2 cos(x). Solution: x2 cos(x) 2x sin(x) ⊕ 2 −cos(x) ⊖ 0 −sin(x) ⊕ The anti-derivative is x2 sin(x) + 2x cos(x) −2 sin(x) + C . Ok, we are now ready for more extreme stuff. 3 9 Find the anti-derivative of x7 cos(x). Solution: x7 cos(x) 7x6 sin(x) ⊕ 42x5 −cos(x) ⊖ 120x4 −sin(x) ⊕ 840x3 cos(x) ⊖ 2520x2 sin(x) ⊕ 5040x −cos(x) ⊖ 5040 −sin(x) ⊕ 0 cos(x) ⊖ The anti-derivative is F(x) = x7 sin(x) + 7x6 cos(x) − 42x5 sin(x) − 210x4 cos(x) + 840x3 sin(x) + 2520x2 cos(x) − 5040x sin(x) − 5040 cos(x) + C . Do this without this method and you see the value of the method. 1 2 3: I myself learned the method from the movie ”Stand and Deliver”, where Jaime Escalante of the Garfield High School in LA uses the method. It can be traced down to an article of V.N. Murty. The method realizes in a clever way an iterated integration by parts method: Z fgdx = fg(−1) −f (1)g−2 + f (2)g(−3) −. . . − (−1)n Z f (n+1)g(−n−1) dx which can easily shown to be true by induction and jus-tifies the method: the f function is differentiated again and again and the g function is integrated again and again. You see, where the alternating minus signs come from. You see that we always pair a k’th derivative with a k + 1’th integral and take the sign (−1)k. Coffee or Tea? 1V.N. Murty, Integration by parts, Two-Year College Mathematics Journal 11, 1980, pages 90-94. 2David Horowitz, Tabular Integration by Parts, College Mathematics Journal, 21, 1990, pages 307-311. 3K.W. Folley, integration by parts, American Mathematical Monthly 54, 1947, pages 542-543 4 When doing integration by parts, We want to try first to differentiate Logs, Inverse trig functions, Powers, Trig functions and Exponentials. This can be remembered as LIPTE which is close to ”lipton” (the tea). For coffee lovers, there is an equivalent one: Logs, Inverse trig functions, Algebraic functions, Trig functions and Exponentials which can be remembered as LIATE which is close to ”latte” (the coffee). Whether you prefer to remember it as a ”coffee latte” or a ”lipton tea” is up to you. There is even a better method, the ”opportunistic method”: Just integrate what you can integrate and differentiate the rest. An don’t forget to consider integrating 1, if nothing else works. Homework 1 Integrate R x2 log(x) dx. 2 Integrate R x5 sin(x) dx 3 Find the anti derivative of R x6 exp(x) dx. () 4 Find the anti derivative of R √x log(x) dx. 5 Find the anti derivative of R sin(x) exp(−x) dx. () If you want to go for the record. Lets see who can integrate the largest xn exp(x)! It has to be done by hand, not with a computer algebra system although. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 29: Worksheet Integration by parts 1 Find the anti-derivative of log(2x)√x: 2 Stand and deliver! Find the anti-derivative of x3 sin(2x): x3 sin(2x) ⊕ ⊖ ⊕ ⊖ Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 30: Numerical integration Before covering two more integration techniques, we briefly look at some numerical techniques. There are variations of basic Riemann sums but speed up the computation. Riemann sum with nonuniform spacing A more general Riemann sum is obtained by choosing n points in [a, b] and defining Sn = X f(yj)(xj+1 −xj) = X yj f(yj)∆xj where yj is in (xj, xj+1). This generalization allows to use a small mesh size where the function fluctuates a lot. The sum P f(xj)∆xj is called the left Riemann sum, the sum P f(xj+1)∆xj the right Riemann sum. xk x y xk x y If x0 = a, xn = b and maxj∆xj →0 for n →∞then Sn converges to R b a f(x) dx. 1 If xj −xk = 1/n and zj = xj, then we have the Riemann sum as we defined it earlier. 2 You numerically integrate sin(x) on [0, π/2] with a Riemann sum. What is better, the left Riemann sum or the right Riemann sum? Look also at the interval [π/2, π]? Solution: you see that in the first case, the left Riemann sum is smaller than the actual integral. In the second case, the left Riemann sum is larger than the actual integral. Trapezoid rule The average between the left and right hand Riemann sum is called the Trapezoid rule. Geometrically, it sums up areas of trapezoids instead of rectangles. 1 2 xk x y The Trapezoid rule does not change things much in the case of equal spacing xk = a + (b −a)k/n. 1 2n[f(x0) + f(xn)] + 1 n n−1 X k=1 f(xk) . Simpson rule The Simpson rule computes the sum Sn = 1 6n n X k=1 [f(xk) + 4f(yk) + f(xk+1)] , where yk are the midpoints between xk and xk+1. The Simpson rule is good because it is exact for quadratic functions: you can check for f(x) = ax2 + bx + c that the formula 1 v −u Z v u f(x) dx = [f(u) + 4f((u + v)/2) + f(v)]/6 holds. To prove it just run the following two lines in Mathematica: (== means ”is equal”) ✞ f [ x ] := a xˆ2 + b x + c ; Simplify [ ( f [ u]+ f [ v]+4 f [ ( u+v)/2])/6==Integrate [ f [ x ] ,{ x , u , v }]/( v−u ) ] ✝ ✆ This actually will imply (as you will see in Math 1b) that the numerical integration for functions which are 4 times differentiable gives numerical results which are n−4 close to the actual integral. For 100 division points, this can give accuracy to 10−8 already. There are other variants which are a bit better but need more function values. If xk, yk, zk, xk+1 are equally spaced, then The Simpson 3/8 rule computes 1 8n n X k=1 [f(xk) + 3f(yk) + 3f(zk) + f(xk+1)] . This formula is again exact for quadratic functions: for f(x) = ax2 + bx + c, the formula 1 v −u Z v u f(x) dx = [f(u) + 3f((2u + v)/3) + 3f((u + 2v)/3) + f(v)]/6 holds. If you are interested, run the two Mathematica lines: ✞ f [ x ] := a xˆ2 + b x + c ; L=Integrate [ f [ x ] ,{ x , u , v }]/( v−u ) ; Simplify [ ( f [ u]+ f [ v]+3 f [ ( 2 u+v)/3]+3 f [ ( u+2v)/3])/8==L ] ✝ ✆ This 3/8 method can be slightly better than the first Simpson rule. 3 Mean value method The mean value theorem shows that for xk = k/n, there are points yk ∈[xk, xk+1] such that f(yj) = F ′(yj) = f(xj+1) −f(xj) and so 1 n Pn k=1 f(yk) = F(xn) −F(x0) . This is a version of the fundamental theorem of calculus which is exact in the sense that for every n, this is a correct formula. Lets call yk the Rolle points. The Rolle point is close to the interval midpoint. For any partition xk on [a, b] with x0 = a, xn = b, there is a choice of Rolle points yk ∈[xk, xk+1] such that the Riemann sum P k f(yk)∆(x)k is equal to R b a f(x) dx. For linear functions the Rolle points are the midpoints. In general, the deviation g(t) from the midpoint is small if the interval is [x0 −t, x0 + t]. One can estimate g(t) to be of the order t2 f′′′(x0) 6f′′(x0). We could modify the trapezoid rule and replace the line through the points by a Taylor polynomial. The Rolle point method is useful for functions which can have poles. 1 Monte Carlo Method A powerful integration method is to chose n random points xk in [a, b] and look at the sum divided by n. Because it uses randomness, it is called Monte Carlo method. The Monte Carlo integral is the limit Sn to infinity Sn = 1 n n X k=1 f(xk) , where xk are n random values in [a, b]. The law of large numbers in probability shows that the Monte Carlo integral is equivalent to the Lebesgue integral which is more powerful than the Riemann integral. Monte Carlo integration is interesting especially if the function is complicated. 3 Lets look at the salt and pepper function f(x) =  1 x rational 0 x irrational The Riemann integral with equal spacing k/n is equal to 1 for every n. But this is only because we have evaluated the function at rational points, where it is 1. The Monte Carlo integral gives zero because if we chose a random number in [0, 1] we hit an irrational number with probability 1. 1I have not found the method yet in the literature. I used it myself when working on a problem in probability theory where functions with poles appeared. 4 The Salt and Pepper function and the Boston Salt and Pepper bridge (Anne Heywood). 4 The following two lines evaluate the area of the Man-delbrot fractal using Monte Carlo integration. The function F is equal to 1, if the parameter value c of the quadratic map z →z2+c is in the Mandelbrot set and 0 else. It shoots 100′000 random points and counts what fraction of the square of area 9 is covered by the set. Numerical experiments give values close to the actual value around 1.51.... One could use more points to get more accurate estimates. ✞ F[ c ]:= Block[{ z=c , u=1},Do[ z= N[ zˆ2+c ] ; If [Abs[ z ] >3 ,u=0;z =3] ,{99}];u ] ; M=10ˆ5; Sum[F[−2.5+3 Random[]+ I(−1.5+3 Random[ ] ) ] , {M}]∗(9.0/M) ✝ ✆ Homework 1 Use a computer to generate 20 random numbers xk in [0,1]. Sum up the square x2 k of these numbers and divide by 20. Compare your result with R 1 0 x2 dx. Remark. If using a program, increase the value of n as large as you can. Here is a Mathematica code: ✞ n=20; Sum[ Random[ ] ˆ 2 , { n}]/ n ✝ ✆ Here is an implementation in Perl. Its still possible to cram the code into one line: ✞ #!/ usr/ bin/ p e r l $n=20; $s =0; for ( $i =0; $i<$n ; $i++){$f=rand ( ) ; $s+=$f ∗$f ;} print $s/$n ; ✝ ✆ 2 a) Use the Simpson rule to compute R π 0 sin(x) dx using n = 2 intervals [0, π/2] and [π/2, π]. On each of these intervals [a, b] compute the Simpson sum [f(a) + 4f((a + b)/2) + f(b)]/6 with f(x) = sin(x). Compare with the actual integral. b) Now use the 3/8 Simpson rule to estimate R π 0 f(x) dx using n = 1 intervals [0, π]. Again compare with the actual integral. Instead of adding more numerical methods exercices, we want to practice a bit more integration. The challenge in the following problems is to find out which integration method is bets suited. This is good preparation for the final, where we will not reveal which integration method is the best. 3 Integrate tan(x)/ cos(x) from 0 to π/6. 4 Find the antiderivative of x sin(x) exp(x). 5 Find the antiderivative of x/ sin(x)2. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 30: Worksheet Numerical methods velocity v time t accident Foto taken by Oliver (while be-ing in grad school) from his Paraglider near Lauterbrunnen in Switzerland. 1 A paraglider starts a flight in the mountain. The velocity is given in the above graph. Find out, whether the paraglider lands lower or higher than where it started. Hint To estimate integrals take the average of the number A of squares entirely below the graph and the number B of squares containing part of the region below the graph. The result A + B is a good estimate for the area below the graph. 2 Review: Integrate x1/3 log(x) dx 3 Review: Integrate log(x5)(1/x) dx Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 31: Partial fractions The partial fraction method will be covered in detail follow up calculus courses like Math 1b. Here we just look at some samples to see whats out there. We have learned how to integrate polynomials like x4 + 5x + 3. What about rational functions? We will see here that they are a piece of cake - if you have the right guide of course ... What we know already Lets see what we know already: • We also know that integrating 1/x gives log(x). We can for example integrate Z 1 x −6 dx = log(x −6) + C . • We also have learned how to integrate 1/(1 + x2). It was an important integral: Z 1 1 + x2 dx = arctan(x) + C . Using substitution, we can do more like Z dx 1 + 4x2 = Z du/2 1 + u2 = arctan(u)/2 = arctan(2x)/2 . • We also know how to integrate functions of the type x/(x2 +c) using substitution. We can write u = x2 + c and get du = 2xdx so that Z x x2 + c dx = Z 1 2u du = log(x2 + c) 2 . • Also functions 1/(x + c)2 can be integrated using substitution. With x + c = u we get du = dx and Z 1 (x + c)2 dx = Z 1 u2du = −1 u + C = − 1 x + c + C . The partial fraction method We would love to be able to integrate any rational function f(x) = p(x) q(x) , where p, q are polynomials. This is where partial fractions come in. The idea is to write a rational function as a sum of fractions we know how to integrate. The above examples have shown that we can integrate a/(x + c), (ax + b)/(x2 + c), a/(x + c)2 and cases, which after substitution are of this type. The partial fraction method writes p(x)/q(x) as a sum of functions of the above type which we can integrate. This is an algebra problem. Here is an important special case: 1 2 In order to integrate R 1 (x−a)(x−b) dx, write 1 (x −a)(x −b) = A x −a + B x −b . and solve for A, B. In order to solve for A, B, write the right hand side as one fraction again 1 (x −a)(x −b) = A(x −b) + B(x −a) (x −a)(x −b) . We only need to look at the nominator: 1 = Ax −Ab + Bx −Ba . In order that this is true we must have A + B = 0, Ab −Ba = 1. This allows us to solve for A, B. Examples 1 To integrate R 2 1−x2 dx we can write 2 1 −x2 = 1 1 −x + 1 1 + x and integrate each term Z 2 1 −x2 = log(1 + x) −log(1 −x) . 2 Integrate R 5−2x x2−5x+6 dx. Solution. The denominator is factored as (x −2)(x −3). Write 5 −2x x2 −5x + 6 = A x −3 + B x −2 . Now multiply out and solve for A, B: A(x −2) + B(x −3) = 5 −2x . This gives the equations A + B = −2, −2A −3B = 5. From the first equation we get A = −B −2 and from the second equation we get 2B + 4 −3B = 5 so that B = −1 and so A = −1. We have not obtained 5 −2x x2 −5x + 6 = − 1 x −3 − 1 x −2 and can integrate: Z 5 −2x x2 −5x + 6 dx = −log(x −3) −log(x −2) . Actually, we could have got this one also with substitution. How? 3 Integrate f(x) = R 1 1−4x2 dx. Solution. The denominator is factored as (1 −2x)(1 + 2x). Write A 1 −2x + B 1 + 2x = 1 1 −4x2 . We get A = 1/4 and B = −1/4 and get the integral Z f(x) dx = 1 4 log(1 −2x) −1 4 log(1 + 2x) + C . 3 Hopital’s method There is a fast method to get the coefficients: If a is different from b, then the coefficients A, B in p(x) (x −a)(x −b) = A x −a + B x −b , are A = lim x→a(x −a)f(x) = p(a)/(a −b), B = lim x→b(x −b)f(x) = p(b)/(b −a) . Proof. If we multiply the identity with x −a we get p(x) (x −b) = A + B(x −a) x −b . Now we can take the limit x →a without peril and end up with A = p(a)/(x −b). Cool, isn’t it? This Hopital method can save you a lot of time! Especially when you deal with more factors and where sometimes complicated systems of linear equations would have to be solved. Remember Math is all about elegance and does not use complicated methods if simple ones are available. Here is an example: 4 Find the anti-derivative of f(x) = 2x+3 (x−4)(x+8). Solution. We write 2x + 3 (x −4)(x + 8) = A x −4 + B x + 8 Now A = 2∗4+3 4+8 = 11/12, and B = 2∗(−8)+3 (−8−4) = 13/12. We have 2x + 3 (x −4)(x + 8) = (11/12) x −4 + (13/12) x + 8 . The integral is 11 12 log(x −4) + 13 12 log(x + 8) . Here is an example with three factors: 5 Find the anti-derivative of f(x) = x2+x+1 (x−1)(x−2)(x−3). Solution. We write x2 + x + 1 (x −1)(x −2)(x −3) = A x −1 + B x −2 + C x −3 Now A = 12+1+1 (1−2)(1−3) = 3/2 and B = 22+2+1 (2−1)(2−3) = −7 and C = 32+3+1 (3−1)(3−2) = 13/2. The integral is 3 2 log(x −1) −7 log(x −2) + 13 2 log(x −3) . And because we like it extreme, here is a larger example: 6 Find the anti-derivative of f(x) = 1 x(x −1)(x −2)(x −3)(x −4)(x −5)(x −6)(x −7)(x −8)(x −9) . Ask your friends whether they have done a partial fraction example with 10th degree polynomial in the denominator. I bet they didn’t do any. Since I have never seen such an 4 example in a text book, look at this example as a ”first”: Solution. We write f(x) = A0 x + A1 x −1 + A2 x −2 + A3 x −3 + A4 x −4 + A5 x −5 + A6 x −6 + A7 x −7 + A8 x −8 + A9 x −9 . The constants are A0 = 1 (0 −1)(0 −2)(0 −3)(0 −4)(0 −5)(0 −6)(0 −7)(0 −8)(0 −9) = −1 362880 A1 = 1 (1 −0)(1 −2)(1 −3)(1 −4)(1 −5)(1 −6)(1 −7)(1 −8)(1 −9) = 1 40320 A2 = 1 (2 −0)(2 −1)(2 −3)(2 −4)(2 −5)(2 −6)(2 −7)(2 −8)(2 −9) = −1 10080 A3 = 1 (3 −0)(3 −1)(3 −2)(3 −4)(3 −5)(3 −6)(3 −7)(3 −8)(3 −9) = 1 4320 A4 = 1 (4 −0)(4 −1)(4 −2)(4 −3)(4 −5)(4 −6)(4 −7)(4 −8)(4 −9) = −1 2880 A5 = 1 (5 −0)(5 −1)(5 −2)(5 −3)(5 −4)(5 −6)(5 −7)(5 −8)(5 −9) = 1 2880 A6 = 1 (6 −0)(6 −1)(6 −2)(6 −3)(6 −4)(6 −5)(6 −7)(6 −8)(6 −9) = −1 4320 A7 = 1 (7 −0)(7 −1)(7 −2)(7 −3)(7 −4)(7 −5)(7 −6)(7 −8)(7 −9) = 1 10080 A8 = 1 (8 −0)(8 −1)(8 −2)(8 −3)(8 −4)(8 −5)(8 −6)(8 −7)(8 −9) = −1 40320 A9 = 1 (9 −0)(9 −1)(9 −2)(9 −3)(9 −4)(9 −5)(9 −6)(9 −7)(9 −8) = 1 362880 . The integral is −log(x) 362880 + log(x −1) 40320 −log(x −2) 10080 + log(x −3) 4320 −log(x −4) 2880 + log(x −5) 2880 − log(x −6) 4320 + log(x −7) 10080 −log(x −8) 40320 + log(x −9) 362880 . Homework 1 R 2dx x2−4. 2 R 5dx 4x2+1. 3 R x3−x+1 x2−1 dx. 4 R 3x2 (x2+x+1)(x−1) dx 5 R 1 (x+1)(x−1)(x+7)(x−3) dx. Use Hopitals method of course! Hint for 3). Subtract first a polynomial. Hint for 4). Find the nominator of Ax+B x2+x+1 + C x−1 and set it 3x2. To do so, multply out. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 40: Worksheet Partial fractions 1 Integrate 1 1+x. 2 Integrate 9 (x−1)2. 3 Integrate 7 x2+1. 4 Integrate 1 1−x4 Hint: write the last one first in the form A/(x2 −1) + B/(1 + x2) Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 32: Trig substitutions Trig substitution is a special case of substitution, where x is a trigonometric function of u or u is a trigonometric function of x. Also this topic is covered more in follow up courses like Math 1b. This lecture allows us to practice more the substitution method. Here is an important example: 1 The area of a half circle of radius 1 is given by the integral Z 1 −1 √ 1 −x2 dx . Solution. Write x = sin(u) so that cos(u) = √ 1 −x2. dx = cos(u)du. We have sin(−π/2) = −1 and sin(π/2) = 1 the answer is Z π/2 −π/2 cos(u) cos(u)du = Z π/2 −π/2 (1 + cos(2u))/2 = π 2 . Lets generalize this a bit and do the same computation for a general radius r: 2 Compute the area of a half disc of radius r which is given by the integral Z r −r √ r2 −x2 dx . Solution. Write x = r sin(u) so that r cos(u) = √ r2 −x2 and dx = r cos(u) du and r sin(−π/2) = −r and and r sin(π/2) = r. The answer is Z π/2 −π/2 r2 cos2(u) du = r2π/2 . Here is an example, we know how to integrate 3 Find the integral Z dx √ 1 −x2 . We know the answer is arcsin(x). How can we do that without knowing? Solution. We can do it also with a trig substitution. Try x = sin(u) to get dx = cos(u) du and so Z cos(u) du cos(u) = u = arcsin(x) + C . 1 2 Here is an example, where tan(u) is the right substitution. You have to be told that first. It is hard to come up with the idea: 4 Find the following integral: Z dx x2√ 1 + x2 by using the substitution x = tan(u). Solution. Then 1 + x2 = 1/ cos2(u) and dx = du/ cos2(u). We get Z du cos2(u) tan2(u)(1/ cos(u)) = Z cos(u) sin2(u) du = −1/ sin(u) = −1/ sin(arctan(x)) . Trig substitution is based on the trig identity : cos2(u) + sin2(u) = 1 Depending on whether you divide this by sin2(u) or cos2(u) we get 1 + tan2(u) = 1/ cos2(u), 1 + cot2(u) = 1/ sin2(u) These identities are worth remembering. Lets look at more examples: 5 Evaluate the following integral Z x2/ √ 1 −x2 dx . Solution: Substitute x = cos(u), dx = −sin(u) du and get Z −cos2(u) sin(u) sin(u)du = − Z cos2(u) du = −u 2 −sin(2u) 4 +C = −arcsin(x) 2 + sin(2 arcsin(x)) 4 +C . 6 Evaluate the integral Z dx (1 + x2)2 . Solution: we make the substitution x = tan(u), dx = du/(cos2(u)). Since 1 + x2 = cos−2(u) we have Z dx (1 + x2)2 = Z cos2(u) du = (u/2) + sin(2u) 4 + C = arctan(u) 2 + sin(2 arctan(u) 4 + C . Here comes an other prototype problem: 7 Find the anti derivative of 1/ sin(x). Solution: We use the substitution u = tan(x/2) which gives x = 2 arctan(u), dx = 2du/(1 + u2). Because 1 + u2 = 1/ cos2(x/2) we have 2u 1 + u2 = 2 tan(x/2) cos2(x/2) = 2 sin(x/2) cos(x/2) = sin(x) . Plug this into the integral Z 1 sin(x) dx = Z 1 + u2 2u 2du 1 + u2 = Z 1 u du = log(u) + C = log(tan(x 2)) + C . Unlike before, where x is a trig function of u, now u is a trig function of x. This example shows that the substitution u = tan(x/2) is magic. Because of the following identities 3 u = tan(x/2) dx = 2du (1+u2) sin(x) = 2u 1+u2 cos(x) = 1−u2 1+u2 it allows us to reduce any rational function involving trig functions to rational functions. Any function p(x)/q(x) where p, q are trigonometric polynomials can be integrated using elementary functions. It is usually a lot of work but here is an example: 8 To find the integral Z cos(x) + tan(x) sin(x) + cot(x) dx for example, we replace dx, sin(x), cos(x), tan(x) = sin(x)/ cos(x), cot(x) = cos(x)/ sin(x) with the above formulas we get a rational expression which involves u only This gives us an integral R p(u)/q(u) du with polynomials p, q. In our case, this would simplify to Z 2u (u4 + 2u3 −2u2 + 2u + 1) (u −1)(u + 1) (u2 + 1) (u4 −4u2 −1)du The method of partial fractions provides us then with the solution. 4 Homework 1 Find the antiderivative: Z √ 1 −4x2 dx . 2 Find the antiderivative: Z (1 −x2)3/2 dx . 3 Find the antiderivative: Z √ 1 −x2 x2 dx . 4 Integrate Z 1 1 + sin(x) using the substitution x = tan(u). Hint. Look at the example in this handout. 5 Compute Z dx cos(x) using the substitution u = tan(x/2). Hint. Look at the example in this handout text and use the identity (1 −u2)/(1 + u2) = cos(x). Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 32: Worksheet Trig Substitutions 1 Integrate √ 1 + x2. Hint. Use x = tan(u). 2 Integrate √ 1 −x2. Hint. Use x = cos(u). 3 Integrate √ x2 −1. Hint. Use x = 1/ cos(u). 4 Integrate arccos(x) 1−x2 . Hint. Use x = cos(u). Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 33: Calculus and Music A music piece is a function Calculus plays a role in music because every music piece just is a function. If you have a loudspeaker with a membrane at position f(t) at time t, then you can listen to the music. The pressure variations in the air are sound waves which reach your ear, where your eardrum oscillates with the function f(t −T) + g(t) where g(t) is background noise and T is a time delay for the sound reach your ear. Plotting and playing works the same way. In Mathematica, we can play a function with ✞ Play [ Sin [2 Pi 1000 x ˆ2] ,{ x , 0 , 1 0 } ] ✝ ✆ This function contains all the information about the music piece. A music ”.WAV” file contains sampled values of the function. A sample rate of 44’100 per second is usual. In .MP3 files essential values are encoded in a compressed way. We take this lecture as an opportunity to review some facts about functions. We especially see that log, exp and trigonometric functions play an important role in music. The wave form and hull A periodic signal is the building block of sound. Assume g(x) is a 2π periodic function, we can generate a sound of 440 Hertz when playing the function f(x) = g(440 · 2πx). If the function does not have a smaller period, then we hear the A tone with 440 Hertz. A periodic function g is called a wave form. 0.2 0.4 0.6 0.8 1.0 -1.0 -0.5 0.5 1.0 0.2 0.4 0.6 0.8 1.0 -1.0 -0.5 0.5 1.0 0.5 1.0 1.5 2.0 -0.4 -0.2 0.2 0.4 0.2 0.4 0.6 0.8 1.0 -3 -2 -1 1 2 3 The wave form makes up the timbre of a sound which allows to model music instruments with macroscopic terms like ”attack, vibrato, coloration, noise, echo, reverbation” and other character-istics. The upper hull function is defined as the interpolation of successive local maxima of f. The lower hull function is the interpolation of the local minima. For the function f(x) = sin(100x) for example, the upper hull function is g(x) = 1 and the lower hull function is g(x) = −1. For f(x) = sin(x) sin(100x) the upper hull function is approximately g(x) = | sin(x)| and the lower hull function is approximately g(x) = −| sin(x)|. 1 2 1 2 3 4 5 -1.0 -0.5 0.5 1.0 We can not hear the actual function because the function changes too fast that we can no-tice individual vibrations. But we can hear the hull function. Simplest examples are change of dynamics in music like creshendi or diminuendi or a vibrato. We can gener-ate a beautiful hull by playing two frequen-cies which are close. You hear interference. The scale Western music uses a discrete set of frequencies. This scale is based on the exponential function. The frequency f is an exponential function of the scale s. On the other hand, if the frequency is known then the scale number is a logarithm. The Midi numbering of musical notes is s = 69 + 12 · log2(f/440) 1 What is the frequency of the Midi tone 100? Solution. We have to solve the above equation for f and get the piano scale function f(s) = 440 · 2(s−69)/12 . Evaluated at 100 we get 2637.02 Hz. The piano scale function f(s) = 440 · 2(s−69)/12 . is an exponential function f(s) = beas which satisfies f(s + 12) = 2f(s). 2 Find the discrete derivative Df(x) = f(x+1)−f(x) of the Piano scale function. Solution: The function is of the form f(x) = A2ax. We have f(x+1) = 2af and so Df(x) = (2a−1)f with a = 1/12. Lets get reminded that such discrete relations lead to the important property d dx exp(ax) = a exp(x) for the exponential function. ✞ midifrequency [ m ] := N[440 2ˆ((m −69)/12)] ✝ ✆ The classical piano covers the 88 Midi tone scale from 21 to 108. The lowest frequency is 27.5Hz, the sub-contra-octave A, the highest 4186.01Hz, the 5-line octave C. 40 60 80 100 1000 2000 3000 4000 3 Here are some mathematical operations which one can do with a piece of music. We will demonstrate some during class. Decomposition in overtones: low and high pass filter It turns out that every wave form can be written as a sum of sin and cos functions. Our ear does this Fourier decomposition automatically. We can here melodies. Here is an example of a decom-position: f(x) = sin(x) + sin(2x)/2 + sin(3x)/3 + sin(4x)/4 + sin(5x)/5. With infinitely many terms, one can also discribe discontinuous functions. Filtering and tuning: pitch and autotune An other advantage of a decomposition of a function into basic building blocks is that one can leave out frequencies which are not good. Examples are low pass or high pass filters. A popular filter is autotune which does not filter but moves the frequencies around so that you can no more sing wrong. If 440 Herz (A) and 523.2 Herz (C) for example were the only allowed frequencies, the filter would change a function f(x) = sin(2π441x) + 4 cos(2π521x) to g(x) = sin(2π440x) + 4 cos(2π523.2x). This filtering is done on the wave form scale. Mixing different functions: rip and remix If f and g are two functions which repre-sent songs, we can look at (f +g)/2 which is the average of the two songs. In real life this is done using tracks. Different instruments can be recorded independently for example and then mixed together. One can for example get guitare g(t), voice v(t) and piano p(t) and form f(t) = ag(t) + bv(t) + c(p(t), where the constants a, b, c are chosen. Differentiate functions: reverb and echo If f is a song and h is some time interval, we can look at g(x) = Df(x) = [f(x + h) −f(x)]/h. Such a differentiation is easy to achieve with a real song. It turns out that for small h, like of order of h = 1/1000, the song does not change much. The reason is that a frequency sin(kx) or hearing the derivative cos(kx) produces the same song. However, if we allow h to be larger, then a reverb or echo effect is produced. Other relations with math Symmetries. Symmetries play an important role in art and science. In geometry we know rotational, translational symmetries or reflection symmetries. Like in geometry, sym-metries play a role both in Calculus as well as in Music. We see some examples in the presentation. Mathematics and music have a lot of overlap. Besides wave form analysis and music ma-nipulation operations and symmetry, there are encoding and compression problems, Diophatine problems like how good frequency rations are approximated by rationals: Why is the chromatic scale based on the twelfth root of 2 so good? Indian music for example uses microtones and a scale of 22. The 12 tone scale is good because many powers 2k/12 are close to rational numbers. I once defined the ”scale fitness” function M(n) = n X k=1 minp,q|2k/n −p q|G(p, q) which is a measure on how good a music scale is. It uses Euler’s gradus suavis (=”degree of pleasure”) function G(n, m) of a fraction n/m which is G(n, m) = 1+E(nm/gcd(n, m)), where the Euler gradus function E(n) = P p|n e(p)(p−1) and p runs over all prime factors p of n and e(p) is the multiplicity. The picture to the left shows Euler’s function G(n, m), the right hand side the scale fitness function in dependence on n. You see that n = 12 is clearly the winner. This analysis could be refined to include scales like Stockhausens 5k/25 4 scale. You can listen to the Stockhausen’s scale with f(t) = sin(2πt100·5[t]/25), where [t] is the largest integer smaller than t. Our familiar 12-tone scale can be admired by listening to f(t) = sin(2πt100 · 2[t]/12). 5 10 15 20 25 30 0.01 0.02 0.03 0.04 0.05 3 The perfect fifth 3/2 has the gradus suavis 1 + E(6) = 1 + 2 = 3 which is the same than the perfect fourth 4/3 for which 1 + E(12) = 1 + (2 −1)(3 −1). You can listen to the perfect fifth f(x) = sin(1000x) + sin(1500x) or the perfect fourth sin(1000x) + sin(1333x) and here is a function representing an accord with four notes sin(1000x) + sin(1333x) + sin(1500x) + sin(2000x). Homework 1 Modulation. How do the following function sound? Listen to them for 10 seconds then draw the hull function. a) f(x) = sin(1000x) −sin(1001x) b) f(x) = sin(x) + cos(tan(1000√x)) c) f(x) = √x cos(10000x) d) f(x) = cos(x) sin(e2x)/2 Here is how to play a function with Mathematica. It will play for 9 seconds: ✞ Play [ Cos[ x ] Sin [Exp[2 x ] ] / x , {x , 0 , 10}] ✝ ✆ Hint. You can play functions online with Wolfram Alpha. Here is an example: ✞ play sin (1000 x) ✝ ✆ 2 Amplitude modulation (AM): If you listen to f(x) = sin(x) sin(1000x) you hear an amplitude change. Draw the hull function. How many increase in amplitudes to you hear in 10 seconds? 3 Frequency Modulation (FM): If we play f(x) = x sin(1000 sin(x)), there are points, where the frequency is low. This is a frequency change. Draw the hull function. 4 Smoothness: If we play the function f(x) = tan(sin(3000 sin(x))), the sound sounds pretty nice. If we change that to f(x) = tan(2 sin(3000 sin(x))), the sound is awful. Can you see why? To answer this, you might want to plot a similar function where 3000 is replaced by 3. 5 A mystery sound: How would you describe the sound f(x) = sin(1/ sin(2π3x))? Our ear can not hear frequencies below 20 Hertz. Why can one still hear something? To answer this, you might want to plot the function from x = 0 to x = 10. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 33: Worksheet Calculus in Music 1 How do you think the function f(x) = sin(10000√x) sounds? 2 What about f(x) = sin(10000x2)? 3 And how about f(x) = arctan(x) sin(tan(x)1000x))? 4 And finally sin(x) sin(1000x)? Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 34: Calculus and Statistics In this lecture, we look at an application of calculus to statistics. We have already defined the probability density function f and its anti-derivative, the cumulative distribution function. This lecture is given by Brian Lukoff. This document is what I had prepared for this lecture. Brian’s handout will be the official course note. Functions In statistics, functions appear at many places. First of all for random variables. Then for probabil-ity density functions and cumulative probability density functions. In order to compute quantities like expectation and variance, we have to integrate. The expectation of probability density function f is m = Z ∞ −∞ xf(x) dx . The variance of probability density function f is Z ∞ −∞ (x2 −m)f(x) dx where m is the expectation. The square root of the variance is the standard deviation. The expectation of the normal distribution f(x) = 1 2πσ2e−(x−m)2/(2σ2) dx is equal m. The standard deviation is σ. The expectation of the geometric distribution f(x) = e−axa Z xe−axa dx = 1/a . The variance of the geometric distribution f(x) = e−axa is 1/a2 and the standard deviation 1/a. To see this, compute (remember Tic Tac Toe!) Z x2e−axa dx = 2/a3 . x2 e−ax 2x −e−ax/a ⊕ 2 e−ax/a2 ⊖ 0 e−ax/a3 ⊕ 1 2 More Problems 1 The uniform distribution on [a, b] is the probability density function which is zero for x outside the interval [a, b] and equal to 1/(b−a) for x ∈[a, b]. Find the mean and standard deviation of this density. 2 The Laplace distribution is also called the double exponential distribution. It has the density 1/(2a)e−a|x|. Find the mean and standard deviation of the Laplace distribution. 3 The Logarithmic distribution on [1, 2] has the density C log(x), where C is a constant which makes it a density. What is the constant C? 4 The Rayleigh distribution is the probability density which is 0 for x < 0 and xe−x2/(2s2)/s2 for x > 0). Verify that its mean is s p π/2 and its variance is (4 −π)σ2/2. 5 The Maxwell distribution is the probability density which is zero for x < 0 and 2 2πx2e−x2(/2a2)/a3 for x ≥0. Verify that its mean is 4a/ √ 2π and its variance is a2(3π − 8)/π. Math 1A: introduction to functions and calculus April 20, 2011, Brian Lukoff Lecture 34: Calculus and Statistics A random variable X is a variable that can take one of many values depending on the outcome of some random process. For example, C could be the random variable that represents the number of heads after flipping a coin twice. Then the probability of getting 0 for C is P(C = 0) = 1 4. Likewise, P(C = 1) = 1 2 and P(C = 2) = 1 2. If a random variable X is discrete, taking on integer values, it must always be true that X k P(X = k) = 1; in other words, the sum of the probabilities of all possible outcomes is 1 (100%). The expected value of a random variable X, denoted E[X], is the mean, or our single best guess (“expectation”) for what any given value of X might be. For a discrete random variable X, the expectation is X k k · P(X = k), 1 For our variable C evaluates to 0·P(C = 0)+1·P(C = 1)+2·P(C = 2) = 0· 1 4+1· 1 2+2· 1 4 = 1; in other words, if we flip a coin twice many times, on average we’ll get 1 head for each pair of flips. The variance of a random variable X, denoted Var[X], is a measure of how spread apart a distribution is (how likely we are to get values of X that are far from the mean). It is X k (k −E[X])2P(X = k), which for our variable C evaluates to (0 −1)2P(C = 0) + (1 −1)2P(C = 1) + (2 − 1)2P(C = 2) = 1 4 + 1 4 = 1 2. Note that we are basically just adding up (k −E[X])2 (which is 0 when k is at the mean of X and gets larger when k gets further away from the mean) and weighting it by the probability of having X = k. Now let’s consider a continuous random variable X that could take on any real number. 2 As an example, H might be the height, in inches, of a randomly chosen Harvard student. Note that since H is continuous, P(H = h) = 0 for any particular height h, so it makes more sense to talk about P(h ≤H < h + ǫ) for some small ǫ > 0. For a random variable, the probability density function for that random variable is a function f(x) such that P(a ≤X ≤b) = Z b a f(x) dx. You can probably see where we are going with this: instead of the sum of all prob-abilities being 1, it will instead be the case that Z ∞ −∞ f(x) dx = 1. 1 2 Similarly, the expected value of a continuous distribution X is E[X] = Z ∞ −∞ xf(x) dx and the variance is Var[X] = Z ∞ −∞ (x −E[X])2f(x) dx. 1 Our height distribution H would have what is called a normal probability density func-tion. (Many natural quantities follow a normal distribution.) The normal probability density function is f(x) = 1 σ √ 2π e−(x−µ)2/2σ2. If X has a normal distribution, then E[X] = Z ∞ −∞ xf(x) dx = µ and Var[X] = Z ∞ −∞ (x −µ)2f(x) dx = σ2. 2 Not every distribution is normal, though. For example, incomes are not normally distributed: most people have relatively moderate incomes, but no one has a negative income and there are a few people that have very high incomes. Some people (see below) argue that income follows an exponential distribution, a distribution with probability distribution function f(x) = λe−λx (where λ > 0 is a constant). Exponential distributions have mean E[X] = Z ∞ 0 xλe−λx dx = 1 λ and variance Var[X] = Z ∞ 0  x −1 λ 2 λe−λx dx = 1 λ2. Another quantity of interest is the cumulative distribution function, which is F(x) = P(X ≤x) = Z x 0 f(x) dx. For our income function, the mean household income in the US in 1997 is a random variable I that is exponentially distributed with mean $35,2001, so λ = 1/35200. The probability that a randomly selected person makes $100,000 or less is P(I ≤100000) = Z 100000 0 1 35200e−t/35200 dt = 1 −e−100000/35200 = 94%. 3 The probability density function f(x) = (a −1)/xa represents a power law distribution, where a > 1 is a constant parameter that changes the shape of the distribution. 1Dragulescu & Yakovenko (2001). Evidence for the exponential distribution of income in the USA. The Euro-pean Physical Journal B, 20, 585-559. 3 Homework 1 The uniform distribution on [a, b] is a distribution where any real number between a and b is equally likely to occur. The probability density function is f(x) = 1/(b −a) for a ≤x ≤b and 0 elsewhere. Verify that f(x) is a valid probability density function (i.e., check that it integrates to 1). 2 Find the mean of the uniform distribution on [a, b]. 3 Explain why the mean you found in problem 2 makes sense intuitively. 4 The Cauchy distribution is important in physics. It has a probability density function of f(x) = 1 π b (x −m)2 + b2. Verify that f(x) is a valid probability density function. 5 Find the cumulative distribution function F(x) for the Cauchy distribution. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 34: Calculus and Statistics In this lecture, we look at an application of calculus to statistics. We have already defined the probability density function f and its anti-derivative, the cumulative distribution function. This lecture is given by Brian Lukoff. This document is what I had prepared for this lecture. Brian’s handout will be the official course note. Functions In statistics, functions appear at many places. First of all for random variables. Then for probabil-ity density functions and cumulative probability density functions. In order to compute quantities like expectation and variance, we have to integrate. The expectation of probability density function f is m = Z ∞ −∞ xf(x) dx . The variance of probability density function f is Z ∞ −∞ (x2 −m)f(x) dx where m is the expectation. The square root of the variance is the standard deviation. The expectation of the normal distribution f(x) = 1 2πσ2e−(x−m)2/(2σ2) dx is equal m. The standard deviation is σ. The expectation of the geometric distribution f(x) = e−axa Z xe−axa dx = 1/a . The variance of the geometric distribution f(x) = e−axa is 1/a2 and the standard deviation 1/a. To see this, compute (remember Tic Tac Toe!) Z x2e−axa dx = 2/a3 . x2 e−ax 2x −e−ax/a ⊕ 2 e−ax/a2 ⊖ 0 e−ax/a3 ⊕ 1 2 More Problems 1 The uniform distribution on [a, b] is the probability density function which is zero for x outside the interval [a, b] and equal to 1/(b−a) for x ∈[a, b]. Find the mean and standard deviation of this density. 2 The Laplace distribution is also called the double exponential distribution. It has the density 1/(2a)e−a|x|. Find the mean and standard deviation of the Laplace distribution. 3 The Logarithmic distribution on [1, 2] has the density C log(x), where C is a constant which makes it a density. What is the constant C? 4 The Rayleigh distribution is the probability density which is 0 for x < 0 and xe−x2/(2s2)/s2 for x > 0). Verify that its mean is s p π/2 and its variance is (4 −π)σ2/2. 5 The Maxwell distribution is the probability density which is zero for x < 0 and 2 2πx2e−x2(/2a2)/a3 for x ≥0. Verify that its mean is 4a/ √ 2π and its variance is a2(3π − 8)/π. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 35: Calculus and Economics In this lecture we look more at applications of calculus and focus mostly on economics. This is an opportunity to review extrema problems. Marginal and total cost Recall that the marginal cost was defined as the derivative of the total cost. Both, the marginal cost and total cost are functions of the quantity of goods produced. 1 Assume the total cost function is C(x) = 10x + 0.01x2. Find the marginal cost and the place where the total cost is maximal. Solution. Differentiate. 2 You sell spring water. The marginal cost to produce depends on the season and given by f(x) = 10 −10 sin(2x). For which x is the total cost maximal? 3 The following example is adapted from the book ”Dominik Heckner and Tobias Kretschmer: Don’t worry about Micro, 2008”, where the following strawberry story appears: (verbatim citation in italics): Suppose you have all sizes of strawberries, from very large to very small. Each size of strawberry exists twice except for the smallest, of which you only have one. Let us also say that you line these strawberries up from very large to very small, then to very large again. You take one strawberry after another and place them on a scale that sells you the average weight of all strawberries. The first strawberry that you place in the bucket is very large, while every subsequent one will be smaller until you reach the smallest one. Because of the literal weight of the heavier ones, average weight is larger than mar-ginal weight. Average weight still decreases, although less steeply than marginal weight. Once you reach the smallest strawberry, every subsequent strawberry will be larger which means that the rate of decease of the aver-age weight becomes smaller and smaller until eventually, it stands still. At this point the marginal weight is just equal to the average weight. Lets recall that if F(x) is the total cost function in dependence of the quantity x, then F ′ = f is called the marginal cost. The function g(x) = F(x)/x is called the average cost. A point where f = g is called a break even point. 4 If f(x) = 4x3−3x2+1, then F(x) = x4−x3+x and g(x) = x3−x2+1. Find the break even point and the points where the average costs are extremal. Solution: To get the break even point, we solve f −g = 0. We get f −g = x2(3x −4) and see that x = 0 and x = 4/3 are two break even points. The critical point of g are points where g′(x) = 3x2 −4x. They agree: 1 2 0.5 1.0 1.5 -1 1 2 3 The following theorem tells that the marginal cost is equal to the average cost if and only if the average cost has a critical point. Since total costs are typically concave up, we usually have ”break even points are minima for the average cost”. Since the strawberry story illustrates it well, lets call it the ”strawberry theorem”: Strawberry theorem: We have g′(x) = 0 if an only if f = g. Proof. g′ = (F(x)/x)′ = F ′/x −F/x2 = (1/x)(F ′ −F/x) = (1/x)(f −g) . Volume extremization 1 Assume the cost to heat a room is V (x) + A(x) −πL(x) where V is its volume, A is the surface area and L(x) = πx is proportional to length x. A conference center hall is is eighth of a sphere. Its volume, surface area and length are V (x) = 4πx3 3 1 8, A(x) = (4π 8 + 3π 4 )x2, L(x) = πx . The costs are π/6x3 + (3π/4 + 4π/8)x2 −πx. To extremize the cost, we can minimize f(x) = x3/6 + 5x2/4 −x . The minimum is achieved at x = (−5 + √ 3)/2. 2 A cone shaped solar loudspeaker has to be a cone of volume π. For optimal charging features, the sum of vertical and horizontal shadow areas hr + πr2 need to be extremized. Can you get a minimum or maximum? Solution. Lets first compute the volume of a cone 3 with maximal radius r and height h. At height z, the radius is rz/h. At z the surface area is A(z) = π(hz/r)2 so that the volume is V = Z h 0 π(r2z2/h2)dz = πr2h/3 = π . This means h = 3/r2 and hr = 3/r. The cross section is f(r) = πr2 + 3/r. Setting f ′(r) = 0, we get the critical point (3/(2π))1/3. Source: Grady Klein and Yoram Bauman, The Cartoon Introduction to Economics: Volume One Microeconomics, published by Hill and Wang. 4 Homework 1 Verify the Strawberry theorem in the case f(x) = cos(x). 2 The production function in an office gives the production Q(L) in dependence of labor L. Assume Q(L) = 500L3 −3L5. Find L which gives the maximal production. This can be typical: For smaller groups, production usu-ally increases when adding more workforce. After some point, bottle necks occur, not all resources can be used at the same time, management and bureaucracy is added, each individuum has less impact and feels less responsi-ble, meetings slow down production etc. In this range, adding more people will decrease the productivity. 3 Marginal revenue f is the rate of change in total revenue F. As total and marginal cost, these are functions of the cost x. Assume the total revenue is F(x) = −5x−x5 +9x3. Find the point, where the total revenue has a local maximum. 4 To find the line y = mx through the points (3, 4), (6, 3), (2, 5). We have to minimize the function f(m) = (3m −4)2 + (6m −3)2 + (2m −5)2 . 5 For any a we look at the solid obtained by rotating the graph of the function f(r) = a sin(r/a) around the axes over the interval [0, π/a]. For which a is the volume locally maximal? P.S. You can see the graph of the volume V (a) in dependence of a below. There are many local maxima. The problem is to find them. 0.5 1.0 1.5 2.0 1 2 3 4 5 6 7 Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 35: Worksheet Calculus in Economics The fact that extremisation is a big deal in economics is already in the word. We look at more examples. Assume we have a couple of data points and we want to find the best line y = mx through this in the sense that the sum of the squares of the points to the line is minimal. This leads to an extremal problem which is a special case of a data fitting problem. It would be more adequate to fit with lines y = mx + b or more generally with functions but then we have more variables and run into multivariable calculus or linear algebra problems much outside the scope of this course. But if we have only one parameter, we get a single variable calculus problem. 1 Find the best line y = mx through the points (1, 1), (3, 2), (2, 5). We have to minimize the function. f(m) = (m −1)2 + (3m −2)2 + (2m −5)2 Find the minimum. Solution. 17/14 Lets take a different set of data points and look at the problem to fit functions of the form y = x + b. 2 Find the best line y = x+b through the points (1, 2), (2, 5), (−1, 2), (4 Source: ”Dominik Heckner and Tobias Kretschmer: Don’t worry about Micro, 2008”, pages 271-274. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 36: Artificial intelligence Today, we study the intriguing question in AI: What does it take to build an artificial calculus teacher? Machines assist us already in many domains: heavy work is done by machines and robots, accounting by computers and fighting by drones. Lawyers and doctors are assisted by artificial intelligence. There is no reason why teaching is different. The web has become a ”gigantic brain” to which virtually any question can be asked or googled: ”Dr Know” in Spielberg’s movie ”AI” is humbled: enter symptoms for an illness and get a diagnosis, enter a legal question and find previous cases. Enter a calculus problem and get an answer. Building an artificial calculus teacher involves calculus itself: such a bot must connect dots on various levels: understand questions, read and grade papers and exams, write good and original exam questions, know about learning and pedagogy. Ideally, it should also have ”ideas” like to ”make a lecture on artificial intelligence”. But first of all, our AI friend needs to know calculus and be able to generate and solve calculus problems. 1 Generating calculus problems Having been involved in a linear algebra book project once, helping to generating solutions to problems, I know that some calculus books are written with help of computer algebra systems. They generate problems and solutions. This applies mostly to drill problems. In order to generate problems, we first must build random functions. Our AI engine ”sofia” knew how to generate random problems with solutions. Random functions are involved when asked ”give me an example of a function”. This is easy: the system would generate functions of reasonable complexity: Call the 10 functions {sin, cos, log, exp, tan, sqt, pow, inv, sca, tra } basic functions. Here sqt(x) = √x and inv(x) = 1/xk for a random integer k between −1 and −3, pow(x) = xk for a random integer k between 2 and 5. sca(x) = kx is a scalar multiplication for a random nonzero integer k between −3 and 3 and tra(x) = x+k translates for a random integer k between −4 and 4. Second, we use addition, subtraction multiplication, division and composition to build more com-plicated functions: A basic operation is an operation from the list {f ◦g, f + g, f ∗g, f/g, f −g }. The operation xy is not included because it is equivalent to exp(x log(y)) = exp ◦(x · log). We can now build functions of various complexities: 1In the academic year of 2003/2004, thanks to a grant from the Harvard Provost, I could work with under-graduates Johnny Carlsson, Andrew Chi and Mark Lezama on a ”calculus chat bot”. We spent a couple of hours per week to enter mathematics and general knowledge, build interfaces to various computer algebra systems like Pari, Mathematica, Macsyma and build a web interface. We fed our knowledge to already known chat bots and newly built ones and even had various bots chat with each other. We conceptionally explored the question of automated learning of the bots from the conversations as well as to add context to the conversation, since bots needs to remember previous topics mentioned to understand some questions. We learned how immense the task is. In the mean time it has become business. Companies like Wolfram research have teams of mathematicians and computer scientists working on content for the ”Wolfram alpha” engine. Having recently seen a group at work here in Cambridge on Mass Av, I guess they generate probably in one day as much content as our Sofia group could do in a week for our ”pet project”. 1 2 A random function of complexity n is obtained by taking n random basic functions f1, . . . , fn, and n random basic operators ⊕1, . . . , ⊕n and forming fn ⊕n fn−1 ⊕n−1 · · · ⊕2 f1 ⊕1 f0 where f0(x) = x and where we start forming the function from the right. 1 Visitor: ”Give me an easy function”: Sofia looks for a function of complexity one: like x tan(x), or x + log(x), or −3x2, or x/(x −3). 2 Visitor: ”Give me a function”: Sofia returns a random function of complexity two: x sin(x) −tan(x), or −e √x + √x or x sin(x)/ log(x) or tan(x)/x4. 3 Visitor: ”Give me a difficult function”: Sofia builds a random function of complexity four like x4e−cos(x) cos(x)+tan(x), or x−√x−ex+log(x)+cos(x), or (1+x)(x cot(x)−log(x))/x2, or (−x + sin(x + 3) −3) csc(x) Now, we can build a random calculus problem. To give you an idea, here are some templates for integration problems: A random integration problem of complexity n is a sentence from the sentence list { ”Integrate f(x) = F(x)”, ”Find the anti derivative of F(x)”, ”What is the integral of f(x) = F(x)?”, ”You know the derivative of a function is f ′(x) = F(x). Find f(x).” }, where F is a random function of complexity n. 4 Visitor ”Give me a differentiation problem”. Sofia: Differentiate f(x) = x sin(x) − 1 x2. The answer is 2 x3 + sin(x) + x cos(x). 5 Visitor: ”Give me a difficult integration problem”. Sofia: Find f if f ′(x) = 1 x + 3 sin2(x) + sin(sin(x))  cos(x). The answer is log(x) + sin3(x) −cos(sin(x)). 6 Visitor: ”Give me an easy extremization problem”. Sofia: Find the extrema of f(x) = x/ log(x). The answer is x = e. 7 Visitor: Give me an extremization problem”. Sofia: Find the maxima and minima of f(x) = x −x4 + log(x). The extrema are − q9 + √ 3153 2/3 −8 3 √ 6 + v u u t8 3 √ 6 − 9 + √ 3153 2/3 1 + 6 r 2 9+ √ 3153−8 3 q 6(9+ √ 3153) ! 225/6 3 √ 3 6 p 9 + √ 3153 . The last example shows the perils of random generation. Even so the function had decent com-plexity, the solution was difficult. Solutions can even be transcendental. This is not a big deal: just generate a new problem. By the way, all the above problems and solutions have been gen-erated by Sofia. The dirty secret of calculus books is that there are maybe a thousand different type of questions which are usually asked. This is a reason why textbooks have become boring clones of each other and companies like ”Aleks”, ”demidec” etc exist which constantly mine the web and course sites like this and homework databases like ”webwork” which contain thousands of pre-compiled problems in which randomness is already built in. Automated problem generation is the ”fast food” of teaching and usually not healthy. But like ”fast food” has evolved, we can expect more and more computer assisting in calculus teaching. Be assured that for this course, the problems have been written by hand (I sometimes use Mathe-matica to see whether answers are reasonable). Handmade problems can sometimes a bit ”rough” but hopefully some were more interesting. I feel that it is not fair to feed computer generated problems to humans. It is possible to write a program giving an answer to ”Write me a final exam”, but the exam would be uninspiring. 3 Corner detection How do we detect corners in pictures? This is necessary to understand pictures, drawings. It might also be needed to see whether a given function is reasonably shaped. There should not be too many ”wiggles” for example. There are various techniques to measure that. One of the best methods in computer vision uses the notion of curvature: Given a function f(x), define the curvature as k(x) = f ′′(x) (1 + f ′(x)2)3/2 . Is is a measure on how much the curve is bent at the point (x, f(x)). Positive curvature means the curve is concave up, otherwise concave down. 8 For a quadratic function f(x) = x2, we have κ(x) = 1/(1 + x2). We see that the curvature is maximal at the lowest part of the parabola. 9 For the function f(x) = √ 1 −x2, we have f ′(x) = −2x/ √ 1 −x2 and f ′′(x) = −(1−x2)−3/2. We have (1 + f ′(x)2) = 1/(1 −x2) and k(x) = −1. 10 Problem: Find the curvature for the graph of f(x) = x5/5 −x. Where is the curvature maximal? -2 -1 1 2 1 2 3 4 -1.0 -0.5 0.5 1.0 -1.0 -0.5 0.5 1.0 -1.0 -0.5 0.5 1.0 -6 -4 -2 2 4 6 Connecting the dots We want to connect points P1, . . . , Pn by a smooth graph. This ”connecting the dots” problem is quite frequent. Our brain does this automatically. We need to see a few glances to ”see” the motion of an object and predict where it will end. We need to connect dots if we drive a car, if we interpret a picture etc. On a more abstract level, we need to connect dots in the landscape of ideas whenever we solve a problem. We want to go from A to B and need to construct intermediate steps. Here is a simple method found by G. Chaikin in 1974 2 which generates a smooth curve through a few points. Given a sequence of n points P1, ..., Pn define a new sequence of 2n −2 points R2, ..., R2n−1 by R2i = 3 4Pi + 1 4Pi+1, R2i+1 = 1 4Pi + 3 4Pi+1 for i = 1, ..., n −1. 2G. Chaikin, An algorithm for high speed curve generation. Computer Graphics and Image Processing 3 (1974), 346-349. 4 One such a step defines a Chaikin step. The limiting curve is called the Chaikin curve defined by the original points. The picture should explain how we get the new points from the old ones: divide each segment into 4 pieces and use the two outer points to get new points. The Chaikin steps produce a smooth curve approximating a given set of points. The pictures show curves in two and three dimensions after applying the method a few times. The method can be used for example to study the complexity of random knots. To answer the question stated initially: like artists have become better using computers it could well be that AI will assist teachers in the future and help them to be more efficient. In any way, the AI dragon breathing down our necks will force us all to stay creative. Homework 1 A calculus bot wants to build a differentiation problem by combining log and sin and exp. Differentiate all of the 6 combinations log(sin(exp(x))), log(exp(sin(x))),exp(log(sin(x))), exp(sin(log(x))), sin(log(exp(x))) and sin(exp(log(x))). 2 Four of the 6 combinations of log and sin and exp can be integrated as elementary functions. Do these integrals. 3 Find the curvature of the sin curve at x = 0, x = π/2 and x = 3π/2. 4 Draw the points (0, sin(0)), (π/2, sin(π/2)), (π, sin(π), (3π/2, sin(3π/2)), (2π, sin(2π)) and connect them with lines. Now do Chaikin iteration for at least 2 generations on paper. 5 Answer each of the following 5 human questions in one sentence: a) What is calculus for you? b) What is the nicest application of calculus? c) Who invented calculus and why? d) What is the fundamental theorem and why is it useful? Math 1A: introduction to functions and calculus Sofia, 2011 Lecture 36: Worksheet This worksheet was authored by Sofia 1, an artificial intelligence cal-culus teacher and student! The bot could also learn evenso only in a primitive way. It had to be told ”learn: ...”. This entire LaTeX file was generated automatically, (except for this introduction section which has, (thanks to this parenthesis) become self-aware and so artificially intelligent.) Derivatives Differentiate the following functions: Level 1 1 a) f(x) = x tan(x) b) f(x) = x + tan(x) c) f(x) = x log(x) d) f(x) = e−xx e) f(x) = cos(x) 1Written in the academic year 2003/2004, thanks to a grant from the Harvard Provost together with Johnny Carlsson, Andrew Chi and Mark Lezama. Sofia was a chat bot which would use computer algebra systems to solve calculus problems while chatting, similar to Wolfram Alpha now. The later is of course much more sophisticated. Ours was maybe a 25 week 4 people 15 hour = half a person-year project Integrals Integrate the following functions: Level 1 1 a) f(x) = 1 x2 + 1 b) f(x) = sec2(x) c) f(x) = 1 −sin(x) d) f(x) = sec2(x) + 1 e) f(x) = 1 2√x Derivatives Differentiate the following functions: Level 2 1 a) f(x) = x3/2 sec(x) b) f(x) = e−x(x + sin(x)) c) f(x) = 0 d) f(x) = e−x√x e) f(x) = ex log(ex) Integrals Integrate the following functions: Level 2 1 a) f(x) = x3−3 x4 b) f(x) = 3√x−3 2 c) f(x) = ex(cos(x) −sin(x)) d) f(x) = x −x tan2(x) −tan(x) e) f(x) = ex (sin (ex) + ex cos (ex)) Derivatives Differentiate the following functions: Level 3 1 a) f(x) = x2(x −tan(x)) b) f(x) = √x −log(x) + x tan(x) c) f(x) = e−x (x−1)x2 d) f(x) = e−x(x+log(x)) x e) f(x) = x3(x + log(x)) Integrals Integrate the following functions: Level 3 1 a) f(x) = −4x4+1 x b) f(x) = e−x((−x log(x) + log(x) + 1) sin(x) + x log(x) cos(x)) c) f(x) = e−2x(1 −2x) −sec2(x) d) f(x) = ex + sec2(x) e) f(x) = (log(x)−2) cos(√x)−√x log(x) sin (√x) √x log2(x) Derivatives Differentiate the following functions: Level 4 1 a) f(x) = sin3(x)+sin(x)+sin(sin(x)) x+1 b) f(x) =  x3 − √ x3 −x −4  csc(x) c) f(x) =  e−3x −9x  cot(x) d) f(x) = e−x((x −3)(x −2) −cos(x)) e) f(x) = √ ex−3x 3x Integrals Integrate the following functions: Level 4 1 a) f(x) = −√x(x−6)+2(x−2)x sin(x)+2(x−4) cos(x) 2x3 b) f(x) = x4 +(x−2)  4x3 −cos(x) −sec2(x)  −sin(x)−tan(x) c) f(x) = − e 1 x(3 sec2( 1 x3)+(3x+1)x2 tan( 1 x3)) x7 d) f(x) = − 2 tan(log(x))+ sin(log(x)) √ cos(log(x)) 2x e) f(x) = 6 tan5(x) sec2(x) Derivatives Differentiate the following functions: Level 6 1 a) f(x) = 1 x3/2 + (√x+log(√x)) cos (√x) cot(√x) √x+3 b) f(x) = −(x + 3) sin 3 2(x)(tan(sin(x)) −log(tan(sin(x)))) c) f(x) = 1 x2 − 2x( x log(x)−log(x)) x+1 + sin (x) d) f(x) = x5  x4 −x + ex −cos(x)  −x3 e) f(x) = sec(x)  sin6(x) + sin(sin(x)) + tan(sin(x))  −ex Integrals Integrate the following functions: Level 6 1 a) f(x) = sec(x)  sec(x) log(tan(x))  8√ tan(x)−sin √ tan(x)  +3 sin  3√ tan(x)  − 8√ tan (x) log2(ta b) f(x) = 1 2    1 √x − e−x  6 log 4(x)+ 1 (log(x)−2)2  x −2e−x 1 log 3(x) + 1 2(log(x)−2) !     c) f(x) = −2x5 log(x) sec2(x)+2x4(3 log(x)−1) tan(x)+log 2(x) x2 log2(x) d) f(x) = −5x7/2+20x6−24x4+2 2x2 e) f(x) = − 2x2√ cos(x)+x log(x) sin(x)+(6 log(x)−2) cos(x) 2 x4√ cos(x) Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 37: The lighter side of Calculus First some serious final thoughts: The holographic picture Any knowledge can be organized in a holographic way, where the amount of detail is a parameter. A 1 second is version is Calculus is great. Calculus in 10 seconds would be ”Calculus establishes that two operations on functions f are related: first the de-rivative of f which is the rate of change of f second the integral of f which is the area under the graph of f.” In 1 minute , I would say something similar than in the synopsis provided by the Harvard registrar for this single variable calculus course: The development of calculus by Newton and Leibniz is a major achievement of the past millennium. The core of this course introduces differential and integral calculus. Differential calculus studies the ”rate of change” f ′ of a function, integral calculus treats ”accumulation” R x 0 f(x) dx which can be interpreted as ”area under the curve”. The fundamental theorem of calculus links the two: it tells that Z x 0 f ′(t) dt = f(x) −f(0), d dx Z x 0 f(t) dt = f(x) . The subject can be applied to problems from other scientific disciplines like eco-nomics (the strawberry theorem for total, average and marginal costs), reasoning for relating quantities (like estimating the speed of an airplane from angle change, psychology (catastrophes explaining revolutionary changes or flips in human per-ception), geometry (volume and area computation), statistics (distribution and cu-mulative distribution functions) or everyday life (the wobbly table theorem). Here is an attempt to summarize the most important points of this course in 3 minutes : 1 2 1. Calculus relates two fundamental operations, the derivative measuring the rate of change of a function and the derivative which measures the area under the graph. 2. Taking derivatives is done with the chain, product and quotient rules, taking intervals with substitution including trig substitution, integration by parts and par-tial fraction rules. 3. Basic functions are polynomials and exp, log, sin, cos, tan. We can add, subtract, multiply, divide and compose functions, we can differentiate and integrate them. 4. A function is continuous at p if f(x) →f(b) for x →p. It is differentiable at p if (f(x + h) −f(x))/h has a limit for h →0. Limits from the right and left should agree. 5. To extremize a function, we look at points where f ′(x) = 0. If f ′′(x) > 0 we have a local minimum, if f ′′(x) < 0 we have a local maximum. There can be critical points f ′(x) = 0 without being extremum like x3. 6) To relate how different quantities change in time, we differentiate the formula relating the quantities using the chain rule. If there is a third time variable, then this is the story of related rates, if one of the variables is the parameter, then this is implicit differentiation. We will have a 90 minutes review of all the material before the midterm. It would not fit on the 4 pages which I strictly allocated as a handout for each lecture. We have seen a 37+13 hour version of the material in the lectures and problem sessions for this course. It would be possible to teach this course using 100 hours . One could explore the material more from a historical perspectives for example and read original sources. One could do projects, use more computer algebra systems, practice visualization and visualize things. You have studied maybe 300 hours for this course including homework, reading, and discussing the material. Years would be needed to study it more on a research level. New calculus is constantly developed. I myself have been working mostly on more probabilistic versions of calculus which allows to bypass some of the difficulties when discretizing calculus. The loss of symmetries obtained by discretization can be compensated differently. The future of calculus Calculus will undoubtably look different in 50 years. Many changes have already started, not only on the context level, also from outside: Calculus books will be gone, electronic paper which will be almost indistinguishable from real paper has replaced it. Text, computations, graphics are all fluid in that we can at any point adjust the amount of details. Similarly than we can zoom into a map or picture by pinching the screen, we can triple pinch a text or proof or picture. As we do so, more details are added, more steps of a calculation added, more information included into a graph etc. Every picture is interactive can turn in a movie, an animation, parameters can be changed, functions deformed with the finger. Every picture is a little laboratory. Questions can be asked directly to the text and answers provided. The text can at any time be set back to an official textbook version of the course. The teacher has the possibility to set global preferences and toss around topics. Examples, homework problems and exam problems will be adjusted automatically disallowing for example to treat integration by parts before the product rule. Much of this is not science fiction, there are electronic interactive books already now available for tablet computers which have impressive experimental and animation features. Impossible because it is too difficult to achieve? Remember the last lecture 36. We will have AI on our side and much of this grunt work to compress and expand knowledge can be done computer assisted. Calculus courses after 1a 3 To prepare for this course, I set myself the task to formulate the main topic in one short sentence and then single out 4 major goals for the course, then build titles for each lecture etc Here are 4 calculus courses at Harvard drawn out at the level of a ”4 point summary”. At other schools of higher education, there are similar courses. The course 1A from extremization to the fundamental theorem functions polynomials, exp, log, trig functions limits velocity, tangents, infinite limits derivatives product, chain rule with related rates, extremization integrals techniques, area, volume, fundamental theorem The course 1B from series and integration to differential equations integration integration: parts, trig substitution, partial fractions, indefinite series convergence, Power, Taylor and Dirichlet series diffequations separation of variables, systems like exponential and logistic equations systems diffeq equilibria, nullclines, analysis The course 21A geometry, extremization and integral theorems in space geometry analytic geometry of space, geometric objects, distances differentiation curves and surfaces, gradient, curl, divergence integration double and triple integrals, other coordinate systems integral theorems line and flux integrals, Green, Stokes and Gauss The course 21B matrix algebra, eigensystems, dynamical systems and Fourier equations and maps Gauss-Jordan elimination, kernel, image, linear maps matrix algebra determinants, eigenvalues, eigenspaces, diagonalization dynamical systems difference and differential equations with various techniques fourier theory Fourier series and dynamical systems on function spaces There is also a 19a/19b track. The 19a course focuses on models and applications in biology, the 19b course replaces differential equations from 21b with probability theory. The Math 20 course covers linear algebra and multivariable calculus for economists in one semester but covers less material than the 21a/21b track. The lighter side of calculus Sofia, our bot had also to know a lot of jokes, especially about math. Here are some relevant to calculus in some way. I left out the inappropriate ones. 1 Why do you rarely find mathematicians at the beach? Because they use sine and cosine to get a tan. 2 Theorem: The less you know, the more you make. Proof: We know Power = Work/Time. Since Knowledge = Power and Time = Money we know Knowledge = Work/Money. Solve for Money to get Money = Work/Knowledge. If Knowledge goes to zero, money approaches infinity. 3 Why do they never serve beer in a calculus class? Because you can’t drink and derive. 4 Descartes comes to a bar. Barmen: An other beer? Descartes: I think not. And disap-pears. 5 If it’s zero degrees outside today. Tomorrow it will be twice as cold. How cold will it be? 6 There are three types of calculus teachers: those who can count and those who can not. 7 Calculus is like love; a simple idea, but it can be complicated. 4 8 A mathematician and an engineer are on a desert island with two palm trees and coconuts. The engineer climbs up, gets its coconut gets down and eats. The mathematician climbs up the other, gets the coconut, climbs the first tree and deposits it. ”I’ve reduced the problem to a solved one”. 9 Pickup line: You are so x2. Can I be x3/3, the area under your curves? 10 The Evolution of calculus teaching: 1960ies: A peasant sells a bag of potatoes for 10 dollars. His costs are 4/5 of his selling price. What is his profit? 1970ies: A farmer sells a bag of potatoes for 10 dollars. His costs are 4/5 of his selling price, that is, 8 dollars. What is his profit? 1980ies: A farmer exchanges a set P of potatoes with a set M of money. The cardinality of the set M is equal to 10, and each element of M is worth one dollars Draw ten big dots representing the elements of M. The set C of production costs is composed of two big dots less than the set M. Represent C as a subset of M and give the answer to the question: What is the cardinality of the set of profits? 1990ies: A farmer sells a bag of potatoes for 10 dollars. His production costs are 8 dollars, and his profit is 2 dollars. Underline the word ”potatoes” and discuss it with your classmates. 2000ies: A farmer sells a bag of potatoes for 10 dollars. His or her production costs are 0.80 of his or her revenue. On your calculator, graph revenue vs. costs and run the program POTATO to determine the profit. Discuss the result with other students and start blog about other examples in economics. 2010ies: A farmer sells a bag of potatoes for 10 dollars. His costs are 8 dollars. Use the Potato theorem to find the profit. Then watch the wobbling potato movie. 11 Q: What is the first derivative of a cow? A: Prime Rib! 12 Q: What does the zero say to the eight? A: Nice belt! 13 Theorem. A cat has nine tails. Proof. No cat has eight tails. Since one cat has one more tail than no cat, it must have nine tails. 14 Q: How can you tell that a mathematician is extroverted? A: When talking to you, he looks at your shoes instead of at his. 15 Q: What does the little mermaid wear? A: An algae-bra. 16 In a dark, narrow alley, a function and a differential operator meet: ”Get out of my way -or I’ll differentiate you till you’re zero!” ”Try it - I’m ex ...” Same alley, same function, but a different operator: ”Get out of my way - or I’ll differentiate you till you’re zero!” ”Try it - I’m ex...” ”Too bad... I’m d/dy.” 17 Q: How do you make 1 burn? A: Fire differentiation at a log. 18 An investment firm hires. In the last round, a mathematician, an engineer, and a business guy are asked what starting salary expectations they had: mathematician: ”Would 30,000 be too much?” engineer: ”I think 60,000 would be OK.” Finance person: ”What about 300,000?” Officer: ”A mathematician will do the same work for a tenth!” Business guy: ”I thought of 135,000 for me, 135,000 for you and 30,000 for the mathematician to do the work. 19 Theorem. Every natural number is interesting. Proof. Assume there is an uninteresting one. Then there is smallest one. But as the smallest, it is interesting. Contradiction! Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 37: Calculus and the world Last Lecture Calculus and the world There would have more to tell. One could make an entire course filled with applications of calculus. We have seen lectures on music, statistics, economics and computer science. Here are more ideas. It would be nice to have a few weeks to work them out. The number of applications explodes even more when doing multivariable calculus or linear algebra. Calculus and Sports Optimization and analysis of motion. Which path needs least energy? Calculus of motion in various sports. Calculus and Biology Exponential growth and decay. Populations grow exponentially. Ra-dioactive particles decay fast. Calculus and Physics Chaos theory. How far into the future can we predict a system. Take a map an iterate it. Take a calculator and iterate. Calculus and Art We can use functions to generate new art forms using functions. Calculus and Cosmology How did the universe evolve. The Lorentz contraction. Is it realistic that we will ever meet an other civilisation. Calculus and Medicine Catastrophes happen also in our body. An example is the story of ”Period doubling” in the heart. Calculus and Finance The mathematics of Finance is complex and is done with stochastic differential equations, chaos theory and power law heuristics. Calculus and Romance When is the optimal time to marry? If you choose too early, you don’t know what is out there. If you chose too late, you will have to compare with too many previous cases. Calculus and Friendship Book by Strogatz: Calculus and Psychology We have seen the catastrophic change of perception. Psychology needs a lot of statistics. Calculus and Politics Game theory and Equilibria. The calculus of conflict. Calculus and Philosophy Is calculus consistent. Can calculus be built in different ways? What is truth? Can we take limits? Calculus and architecture The topic is much linked that most calculus books feature architecture on their book covers. Calculus and History The calculus wars between Newton and Leibniz. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 38: Review since second midterm Related rates Implicit differentiation and related rates are manifestations of the chain rule. A) related rates: we have an equation F(x, y) = c relating two variables x, y which depend on time t. differentiate the equation with respect to t using the chain rule and solve for y′. B) implicit differentiation: we have an equation F(x, y(x)) = c relating y with x. Differentiate the equation with respect to x using the chain rule and solve for y′. Examples: A) x3 + y3 = 1, x(t) = sin(t), then 3x2x′ + 3y2y′ = 0 so that y′ = −x2x′/y2 = −sin2(t) cos(t)/(1 − sin3(t))1/3. B) Same example but x(t) = x: y′ = −x2/y2 = −sin2(t)/(1 −sin3(t))1/3. Substitution Substitution replaces R f(x) dx with R g(u) du with u = u(x), du = u′(x)dx. Special cases: A) The antiderivative of f(x) = g(u(x))u′(x), is G(u(x)) where G is the anti derivative of g. B) R f(ax + b) dx = F(ax + b)/a where F is the anti derivative of f. Examples: A) R sin(x5)x4 dx = R sin(u) du/5 = −cos(u)/5 + C = −cos(x5)/5 + C. B) R log(5x + 7) dx = R log(u) du/5 = (u log(u) −u)/5 + C = (5x + 7) log(5x + 7) −(5x + 7) + C. Integration by parts A) Direct: Z x sin(x) dx = x (−cos(x)) − Z 1 (−cos(x) ) dx = −x cos(x) + sin(x) + C dx . B) Tic-Tac-Toe: To integrate x2 sin(x) x2 sin(x) 2x −cos(x) ⊕ 2 −sin(x) ⊖ 0 cos(x) ⊕ The anti-derivative is −x2 cos(x) + 2x sin(x) + 2 cos(x) + C . C) Merry go round: Example I = R sin(x)ex dx. Use parts twice and solve for I. 1 2 Partial fractions A) Make a common denominator on the right hand side 1 (x−a)(x−b) = A(x−b)+B(x−a) (x−a)(x−b) . and compare coefficients 1 = Ax −Ab + Bx −Ba to get A + B = 0, Ab −Ba = 1 and solve for A, B. B) If f(x) = p(x)/(x −a)(x −b) with different a, b, the coefficients A, B in p(x) (x−a)(x−b) = A x−a + B x−b can be obtained from A = lim x→a(x −a)f(x) = p(a)/(a −b), B = lim x→b(x −b)f(x) = p(b)/(b −a) . Examples: A) R 1 (x+1)(x+2) dx = R A x+1 dx+ R B x+2 dx. Find A, B by multiplying out and comparing coefficients in the nominator. B) Directly write down A = 1 and B = −1, by plugging in x = −2 after multiplying with x −2. or plugging in x = −1 after multiplying with x −1. Improper integrals A) Integrate over infinite domain. B) Integrate over singularity. Examples: A) R ∞ 0 1/(1 + x2) = arctan(∞) −arctan(0) = π/2 −0 = π/2. B) R 1 0 1/x2/3 dx = (3/1)x1/3|1 0 = 3. Trig substitutions A) In places like √ 1 −x2, replace x by cos(u). B) Use u = tan(x/2), dx = 2du (1+u2) , sin(x) = 2u 1+u2, cos(x) = 1−u2 1+u2 to replace trig functions by polynomials. Examples: A) R 1 −1 √ 1 −x2 dx = R π/2 −π/2 cos(u) cos(u)du = R π/2 −π/2(1 + cos(2u))/2 = π 2. B) R 1 sin(x) dx = R 1+u2 2u 2du 1+u2 = R 1 u du = log(u) + C = log(tan( x 2)) + C. Applications, keywords to know Music: hull function, piano function Economics: average cost, marginal cost and total cost. Strawberry theorem, fit points Computer science: curvature and Chaikin steps Statistics: probability density function, cumulative distribution function, expectation, variance. Geometry: area between two curves, volume of solid Numerical methods: trapezoid rule, Simpson rule, Newton Method Psychology: critical points and Catastrophes. Physics: position, velocity and acceleration. Gastronomy: turn table to prevent wobbling, bottle calibration. Math 1A: introduction to functions and calculus Oliver Knill, 2011 Lecture 39: Checklists Integrals to know well sin(x) cos(x) tan(x) log(x) exp(x) 1/x xn 1/ cos2(x) 1/ sin2(x) 1/(1 + x2) 1/ √ 1 −x2 Apps to know Since there are few questions on what has to be known about applications and definitions (this list only covers application parts): Derivative: Limit of differences Dhf = [f(x + h) −f(x)]/h for h →0 Integral: Limit of Riemann sums Shf = [f(0) + f(h) + ...f(kh)]h. Newton step: T(x) = x −f(x)/f ′(x). Marginal cost: the derivative F ′ of the total cost F. Average cost: F/x where F is the total cost. Velocity: Derivative of the position. Acceleration: Derivative of the velocity. Curvature: f ′′(x)/(1 + f ′(x))3/2. Probability distribution function: nonnegative function with total R f(x)dx = 1. Cumulative distribution function: anti-derivative of the probability distribution function. Expectation: R xf(x) dx, where f is the probability density function. Piano function: frequencies f(k) = 440 · 2k/12 for integer k. Hull function: The interpolation of local maxima. Catastrophe: A parameter c at which a local minimum disappears. 1 2 Not on your fingertips The following concepts have appeared but do not need to be learned by heart: Entropy: − R f(x) log(f(x)) dx. Moment of inertia: R x2f(x) dx. Monte Carlo integration: Sn = 1 n Pn k=1 f(xk) , where xk are random in [a, b]. Weierstrass function: A function which is continuous but nowhere differentiable. Bart Simpson rule: Sn = 1 6n Pn k=1[f(xk) + 4f(yk) + f(xk+1)]. Chaikin step: R2i = 3 4Pi + 1 4Pi+1, R2i+1 = 1 4Pi + 3 4Pi+1. Cocktail party stuff: Eat,integrate and love, the story of exp in practice exam 2. Bottles: How to calibrate bottles. The calibration formula. Sofia: The name of a calculus bot. Wobbly chair: One can turn a chair on any lawn to stop it from wobbling. Warthog: ”Tuk”, the name of the warthog which appears in practice exam 2.
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https://www.nature.com/articles/srep03198
Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript. Advertisement Universal statistics of the knockout tournament Scientific Reports volume 3, Article number: 3198 (2013) Cite this article 3695 Accesses 5 Citations 1 Altmetric Metrics details Subjects Abstract We study statistics of the knockout tournament, where only the winner of a fixture progresses to the next. We assign a real number called competitiveness to each contestant and find that the resulting distribution of prize money follows a power law with an exponent close to unity if the competitiveness is a stable quantity and a decisive factor to win a match. Otherwise, the distribution is found narrow. The existing observation of power law distributions in various kinds of real sports tournaments therefore suggests that the rules of those games are constructed in such a way that it is possible to understand the games in terms of the contestants' inherent characteristics of competitiveness. Similar content being viewed by others Quantifying momentum and influencing factors of tennis players using the XGBoost model Interpersonal strategy for controlling unpredictable opponents in soft tennis Evaluation of the winning technical and tactical actions of the best female table tennis players in the years 1970-2021 Introduction Competition is a ubiquitous form of social interaction for distributing limited resources among a number of individuals, often regarded as the opposite of cooperation. Competition has been a main tenet in economics where a perfectly competitive equilibrium is proven Pareto-efficient as long as there are no externalities and public goods. Moreover, the notion of natural selection in biological evolution is often understood as proving competition ‘natural’. For these reasons, although competition results in growing tension across a society, most people have taken it for granted as an organising principle of our society. Recently, Deng et al.1 claimed universal power-law distributions of scores and prize money by observing various kinds of sports such as tennis, golf, football, badminton and so on. According to their extensive data analysis, the probability to find scores or prize money greater than k always decays as a power law P>(k) ~ k−(γ−1) with an exponential cutoff where the power-law exponent γ − 1 ranges between 0.01 and 0.39 depending on sports. In addition, they presented a knockout-tournament model to explain the observations. This is an intriguing approach since the most organised forms of competition are usually found in sports. It is also popular to run a knockout tournament, consisting of successive rounds where only a winner in each fixture progresses to the next round, because it is an efficient procedure to find who is the best with a small number of fixtures. In other words, Deng et al. hinted a direct connection between the structure of competition and its consequences. Physicists have already recognised sports as a fruitful research field: Statistics of athletic records has been pioneered by Gembris et al.2 and Wergen et al.3, for example and there have been attempts to even predict the limiting performances in the long run4. Sports ranking combinatorics has also been considered by Park and Newman5,6. If we are to understand the dynamics governing high achievements in sports careers, in particular, one famous theory along this direction is called the Matthew “rich get richer” effect7,8,9: It says that a higher position leads to a better chance to progress further in career, resulting in an extremely skewed distribution. The spatial Poisson process to model this effect indeed explains such behaviour with γ ≤ 1, which is found in some empirical data sets. However, we should point out that many factors of competition are hidden in the probability of progress and that the stochastic process is totally indifferent to individual characteristics as written in Ecclesiastes: “the race is not to the swift, but time and chance happenth to them all”. In this work, we instead focus on statistical analysis of a specific system of competition, i.e., the knockout tournament among inhomogeneous participants. Our main point is that a large part of statistics is universal in the sense that it is independent of most details of the game but already determined by the tournament structure. Let us consider a player's number of wins denoted by n, for example. When the tournament has been finished, the distribution of n denoted by P(n) is always an exponentially decreasing function of n. It is a purely geometric property of the tournament tree independent of any details of the game, loosely mapped to the critical percolation on a binary tree10. If the prize money is highly skewed towards the best players, similarly to real sports tournaments, one can assume that the prize money kn after winning n rounds is also an exponential function of n, that is, kn ~ zn (Fig. 1). Combining these two, one finds that the distribution P(k) ~ k−γ with and this mechanism belongs to combination of exponentials according to Newman11. If z gets very large, γ converges to unity, yielding P(k) ~ k−1. As z → 1, on the other hand, γ diverges because P(k) approaches the distribution function of n, which is an exponential function. In fact, if z < 2, the total amount of prize money gets unbounded as the number of contestants grows, which means that the organiser of this tournament has a risk of bankruptcy. This explains why kn has to be such a rapidly increasing function of n and we see that the feasible range of γ is between one and two. Moreover, if there is a typical number of prize winners, z is effectively very large, driving γ to unity. This is a simple prediction for a single tournament. In other words, this analysis corresponds to gathering data of prize money distributed over many tournaments without identifying who was who. The actual statistics collected in this way, however, will not be very interesting to us and it is usually more meaningful to consider individual-based statistics: Even for a team sport, each team may be regarded as an individual. It is notable that Deng et al. resolve this problem by introducing the notion of ranks, belonging to individuals and also by assuming that a player's winning probability against another is a function of their rank difference. Following this approach, we will see how our simple prediction in equation (1) can be reproduced on average in the individual-based statistics. Schematic illustration of a tournament with four contestants A, B, C and D. Contestant B has competitiveness rB and gets prize money kB = z2 because she has defeated A and C. Likewise, C gets kC = z1 because she has won only a single match against D. Results Decisiveness of competitiveness Imagine a tournament with N = 2m contestants to construct a simple binary tree. Each person is assigned a real number r, which we refer as competitiveness instead of a rank and reserve the latter term for denoting an outcome of competition, which may or may not reflect an individual's genuine competitiveness depending on how much luck comes into play. By defining r as a real number, the competitiveness is automatically assumed to be transitive, which means that if contestant A is more competitive than B who is more competitive than C, then A is also more competitive than C. Since we can always rescale the highest competitiveness as unity and the lowest one as null without loss of generality, the real number r belongs to a closed interval from zero to one. Under total uncertainty about the contestants, we may assume as our initial condition that the distribution of r is uniformly random at the starting point. We thus denote the initial probability density distribution of r as p0(r) = 1 with normalisation . Then, we introduce a function f(r, r′) that defines the probability for a contestant with competitiveness r to defeat another with r′. As was done by Deng et al.1, it can be assumed to be a function of x ≡ r − r′ only and it is plausible in such a case that f(x) is a nondecreasing function of x ∈ [−1, 1] with f(x) + f(−x) = 1. In words, the former condition means that a more competitive player has a higher probability to defeat a less competitive player, whereas the latter condition is merely a simple reflection of the trivial fact that one of the two players must win, irrespective of their values of r. Let us check some examples of f(x). Perfect resolution One of the simplest choices is where Θ is the Heaviside step function. This means that the competitiveness decides the outcome deterministically. In Methods, we have derived the following nonlinear recursive relation where pn(r) means the distribution of r after the nth round. With the Heaviside step function, this equation is solvable at any arbitrary n and we obtain with a corresponding cumulative distribution . As explained in Methods, cn(r) is identical to the winning chance for the contestant with r at the (n + 1)th round, denoted by wn(r), when we have chosen the step function in equation (2). We can extract various useful information from this probability density function. For example, the average competitiveness after the nth round is and therefore the width of pn(r) decreases as σ ~ 2−n. A contestant with r passes the nth round but not the next one with probability where we have used wk = ck and the sum over n is normalised to unity for any r between zero and one. The average prize money for this person with r can thus be calculated as As shown in Fig. 2, qn has a peak at and the summations above can be approximated as If kn = zn, it means that in the vicinity of r = 1. Note that we have approximated r as unity at the denominator of equation (8). Therefore, Zipf's plot shows a power law with slope −log2 z, leading to P(k) ~ k−γ with γ = (log2 z)−1 + 1 due to the relationship between Zipf's plot and P(k)12. This exactly coincides with equation (1) derived for a single tournament. We have numerically performed tournaments and the results confirm validity of our analysis as shown in Fig. 3, where the numerical calculations of c5(r) and 〈r〉n agree perfectly with the analytic results. The detailed procedure of our simulation is explained in Methods. Conditional probability to progress only to the nth round for given competitiveness r [see equation (6)]. (a) Probability distribution of r at the 5th round when f(x) is the Heaviside step function, equation (2). The data points are obtained numerically by simulating 104 tournaments with N = 212 and the line shows our analytic prediction in equation (4). (b) Average value of r at the nth round, where the data points are obtained numerically and the line represents equation (5). Imperfect resolution As an opposite extreme case, let us consider a situation where individual competitiveness is totally irrelevant to the outcome of a match and only luck decides. In other words, we assume a constant function f(x) = 1/2. If we start from p0 = 1, the winning chance here is . Note that w0 is not identical to the cumulative distribution any more. The next round has a distribution p1(r) = 2w0(r)p0(r) = 1 and this pattern is repeated all the way leading to pn(r) = 1 for every n. It is also straightforward to obtain the same result by substituting the constant f(x) = 1/2 into the recursive equation (3). The resulting P(k) is just the most likely distribution of the prize money among the N players, so the maximum entropy principle tells us to maximize where the first term is Shannon entropy and μ represents a Lagrangian multiplier for constraining the average prize money. When H is maximised, it does not change under variation in P(k) to the first order and we thus have which leads us to P(k) ~ exp(−k/kc) with a characteristic scale kc. This implies a tendency that P(k) usually exhibits a power law with an exponent close to unity but that randomness makes the tail shorter. Suppose that f(x) has a finite resolving power, quantified by a characteristic width Γ over which f(x) rapidly increases. The Heaviside step function corresponds to a limiting case of Γ → 0. We can predict the followings when Γ is finite but sufficiently small: At the beginning of the competition, the width σ of pn(r) is much greater than Γ, so f(x) effectively serves as a step function. The above analysis shows that σ decreases as 2−n so it becomes comparable with Γ after ν ~ log2(1/Γ) rounds. Thereafter, the decrease of σ slows down. Finally, when after many rounds, the survivors' competitiveness is irrelevant and the outcomes are mostly determined by pure luck. Therefore, a natural guess for P(k) would be with kΓ ~ O(zν) and γ in equation (1). This functional form is confirmed in our numerical simulations (Fig. 4). This distribution can also be derived from the maximum entropy principle as in equation (10) but with an additional constraint on Σk ln k13,14, which corresponds to the total number of fixtures in this context. The above argument can be pursued further by employing the following f(x): where the exponential functions make it possible to explicitly evaluate the integral. Then, the winning chance is given as which approaches c0(r) = r as Γ → 0 and c0(r) = 1/2 as Γ → ∞, as expected. As above, this yields which is normalised to unity as . This result is quite suggestive, because equation (16) modifies equation (4) at n = 1 by adding O(Γ) when and subtracting the same amount when [Fig. 5(a)]. In short, p0(r) becomes flatter when r is close to 0 or 1. If we take one step further, the low-r correction becomes less important and we find where we have left only the dominant correction of O(Γ) [Fig. 5(b)]. For general n, the result up to the correction of O(Γ) is inductively found as This implies that the finite resolution is most noticeable among highly competitive players with , whereas the story looks similar to the case of perfect resolution when (1 − r) is small but still much larger than Γ. Cumulative distribution of prize money, where the horizontal axis is rescaled with respect to the largest value. The data points are obtained numerically by simulating 104 tournaments with N = 212 and z = 2, in ascending order of Γ from below. The straight line shows our analytic prediction for Γ = 0 for comparison. Effects of imperfect resolution. (a) p1(r), the distribution of competitiveness after the first round and (b) p2(r) after the second round. The resolution parameter is the width of f(x), which is set to be Γ = 5% here. For comparison, the dotted lines show the cases for Γ = 0. Stability of competitiveness We have assumed that competitiveness is each individual's inherent characteristic, which changes in a much longer time scale compared to outcomes of competition and we relate the latter to ranks. The idea is that although a contestant's rank fluctuates over tournaments, it will correctly reflect her true competitiveness in the long run. Even if the competitiveness may interact with actual tournament results, it will usually be related to a cumulative measure of performance that mainly reflects low-frequency, i.e., long-term behaviour. For example, we have calculated the Kendall tau rank correlation coefficient15, denoted by τ, to see how the accumulated amounts of prize money change their relative positions between two successive tournaments (Fig. 6). If a certain pair of contestants keep their relative positions, they are said to be concordant and discordant otherwise. The coefficient τ is defined as the number of concordant pairs minus that of discordant pairs, divided by the total number of possible pairs. Beginning with the same initial amount of money for every contestant, which is set to zero, we run fifty tournaments in a row, accumulating the prize money for each individual. A contestant's accumulated money from a series of tournaments determines her performance in the next tournament in such a way that r = (N − i)/(N − 1) is assigned to the contestant when she has the ith largest accumulated amount. The relative positions of two equal amounts are random. In spite of this variability, the ranks of the accumulated money get stabilised after 20 or 30 tournaments in all the cases considered (Fig. 6) and the resulting P(k) is almost identical to the static-r case for each Γ. Still, one may ask what happens if their time scales approach each other so that a current rank directly affects performance at the next tournament, provided that the tournaments are regular events. Even if an individual's rank fluctuates over time, it might still be possible for this correlation between successive tournaments to reproduce the power-law tail part of P(k). In fact, this question is not really well-posed because a knockout tournament leaves many contestants' ranks undetermined except a few prize winners and this is the fundamental advantage of a knockout tournament. We nevertheless suppose that a player's competitiveness at the next time step is a nondecreasing function of the current performance, say, rt+1 = R(nt), where nt is the number of wins in the tournament at time t and R is a nondecreasing function between zero and one. Since r determines how many rounds the contestant can go through, the distribution of nt+1 is essentially a function of nt. The situation is actually boring because the same contestant wins the first place all the time, but we may exclude this exceptional contestant from our consideration. We begin with noting that any tournament results in a distribution of nt as , which is the initial distribution of the next tournament at time t + 1. The corresponding cumulative portion of contestants with results below nt is thus . As above, if f(x) is the Heaviside step function with f(0) = 1/2, the chance to win the first round for a contestant that passed nt rounds at the previous tournament is . The first term represents the probability to meet an opponent with the same nt and the factor of one half originates from f(0). The distribution of nt at the next round is p1(nt) = 2w0(nt)p0(nt). We can repeat this procedure to obtain a general expression as with g(x) ≡ 2x. By definition, we have If k is not very small, pk(nt) converges to a certain function of y ≡ k − nt with a maximum around y ≈ 0 [Fig. 7(a)]. The conditional probability to reach k and stop there for given nt is found as with [Fig. 7(b)]. We observe that qk(nt) can also be described as a certain function V(y) when . Moreover, we find that for any nt. In other words, the time series {nt ≥ 0} can be roughly described as a biased random walk towards the origin. Since this holds true for anyone, each contestant's average result will be rapidly equalised by the bias so we predict that the probability distribution P(k) will be narrow. This prediction is well substantiated by numerical results shown in Fig. 8, where P>(k) is drawn in a semi-log plot. Therefore, in terms of the time scale of competitiveness, the power-law shape of P(k) is observable when competitiveness changes much more slowly compared to the frequency of tournaments. Behaviour of the Kendall tau rank correlation coefficient for the contestants' performance when the each contestant's cumulative prize money determines her competitiveness. (a) The horizontal axis means the result of a tournament at time t and the vertical axis means probability to find a contestant with nt at the kth round of the next tournament at t + 1. Note the similarity in shape at , which means that pk(nt) ≈ U(y) with y ≡ k − nt. (b) Conditional probability qk(t) also converges to a certain function V(y) (see text). Cumulative distribution of prize money, when each contestant's tournament result at time t determines her competitiveness at t + 1. We have numerically simulating 104 tournaments with N = 212 and z = 2. We have used the Heaviside step function as f(x) and this plot has excluded the one that always wins the first place. Discussion In summary, we have investigated statistics resulting from knockout tournaments. It is basically the rules of the game that define competitiveness, so the distribution of prize money is dependent on how much the rules take individual competitiveness as a decisive and stable factor. But other details of the game are found irrelevant and the statistics is universal in this sense. More specifically, if competitiveness is a static parameter and any tiny difference of it can be distinguished by the rules, the distribution is predicted to take a power-law shape P(k) ~ k−γ with γ close to unity. If the difference is indistinguishable below a certain resolution limit Γ, we find an exponential cutoff at the tail, whose location is a function of Γ. We have also argued that the distribution P(k) becomes narrow again when competitiveness changes with a time scale comparable to the frequency of tournaments. In this respect, the broad distributions observed across many sports suggest that their rules are already stabilised in such a way that one can readily compare contestants' competitiveness in a consistent way over a long time span and that the result of competition sensitively reflects the difference indeed. Since our analysis relates certain internal parameters of a given tournament such as z and Γ to the final distribution of prize money, which is somewhat more easily accessible, it will an interesting question to verify such detailed relationships directly on empirical grounds. Methods Recursive relation for pn(r) In case of perfect resolution, i.e., f(r, r′) = Θ(r − r′), it is straightforward to obtain the winning chance for the contestant with r at the first round of the tournament as where p0(r′) = 1. This happens to be identical to the cumulative distribution c0(r) and it represents the simple fact that the contestant with r should meet an opponent with r′ < r in order to win and progress to the next round. When the first round has been finished, the distribution of their competitiveness is which is again normalised to unity. The factor of two in front is needed because the number of survivors has become one half of N. Note that we have used independence between a player's competitiveness and her opponent's in equation (23), which is the case when the initial condition contains no correlations in competitiveness. As in the first round, the corresponding cumulative distribution, is identical to the winning chance w1(r) at the second round. In the same way, the distribution after the second round is p2(r) = 2w1(r)p1(r) = 4w1(r)w0(r)p0(r) = 4r3 and so on. For general f(r, r′), we can use essentially the same argument to derive the following nonlinear recursive relation: which is explicitly solvable for a few special cases as above. Numerical procedures First, we generate a tournament tree with N = 2m contestants at the terminal nodes and assign to each of them a real random number r inside the unit interval as competitiveness. One may require the minimum and maximum of the random numbers to be strictly zero and one, respectively, but it does not make a visible difference when N is large enough. The resulting uncorrelated random number sequence {r1, r2, …, rN} means absence of a seeding process, so number one and number two seeds may face each other in the first round. Second, when two contestants A and B meet with rA and rB, respectively, we draw a random number ρ ∈ [0, 1) and choose A as the winner of this fixture if ρ < f(rA, rB) and choose B otherwise. This is repeated for every match in this first round and the winner progresses to the parent node. When we have filled all the parent nodes with 2m−1 winners, the second round starts among them in the same way as before. As the tournament proceeds round by round, the number of survivors decreases rapidly until the final winner is left alone after the mth round. Each player defeated at the nth round receives prize money zn−1, whereas the final winner acquires zm. When a tournament is over, we start a new one with randomly shuffling {r1, r2, …, rN} at the terminal nodes, so that the competitiveness is identified as an individual characteristic preserved across the tournaments. We have performed 104 shuffles, hence the same number of tournaments, to obtain statistical averages for each ri with N = 212. References Deng, W., Li, W., Cai, X., Bulou, A. & Wang, Q. A. Universal scaling in sports ranking. New J. Phys. 14, 093038 (2012). Article ADS Google Scholar Gembris, D., Taylor, J. G. & Suter, D. Sports statistics: Trends and random fluctuations in athletics. Nature (London) 417, 506 (2002). Article CAS ADS Google Scholar Wergen, G., Volovik, D., Redner, S. & Krug, J. Rounding effects in record statistics. Phys. Rev. Lett. 109, 164102 (2012). Article CAS ADS Google Scholar Radicci, F. Universality, limits and predictability of gold-medal performances at the Olympic Games. PLoS ONE 7, e40335 (2012). Article ADS Google Scholar Park, J. & Newman, M. E. J. A network-based ranking system for US college football. J. Stat. Mech. P10014 (2005). Park, J. Diagrammatic perturbation methods in networks and sports ranking combinatorics. J. Stat. Mech. P04006 (2010). Petersen, A. M., Jung, W.-S., Yang, J.-S. & Stanley, H. E. Quantitative and empirical demonstration of the Matthew effect in a study of career longevity. Proc. Natl. Acad. Sci. USA 108, 18–23 (2011). Article CAS ADS Google Scholar Petersen, A. M., Penner, O. & Stanley, H. E. Methods for detrending success metrics to account for inflationary and deflationary factors. Eur. Phys. J. B 79, 67–78 (2011). Article ADS Google Scholar Petersen, A. M., Jung, W.-S. & Stanley, H. E. On the distribution of career longevity and the evolution of home-run prowess in professional baseball. EPL 83, 50010 (2008). Article ADS Google Scholar Baek, S. K., Minnhagen, P. & Kim, B. J. Percolation on hyperbolic lattices. Phys. Rev. E 79, 011124 (2009). Article ADS Google Scholar Newman, M. E. J. Power laws, Pareto distributions and Zipf's law. Contemp. Phys. 46, 323–351 (2005). Article ADS Google Scholar Kim, B. J. & Park, S. M. Distribution of Korean family names. Physica A 347, 683–694 (2005). Article ADS Google Scholar Baek, S. K., Bernhardsson, S. & Minnhagen, P. Zipf's law unzipped. New J. Phys. 13, 043004 (2011). Article ADS Google Scholar Visser, M. Zipf's law, power laws and maximum entropy. New J. Phys. 15, 043021 (2013). Article ADS Google Scholar Kendall, M. A new measure of rank correlation. Biometrika 30, 81–89 (1938). Article Google Scholar Download references Acknowledgements We are indebted to Petter Minnhagen for introducing us to this problem. We thank Korea Institute for Advanced Study for providing computing resources (KIAS Center for Advanced Computation, Abacus System) for this work. B.J.K. was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (No. 2011-0015731). Author information Authors and Affiliations Department of Physics, Pukyong National University, 608-737, Busan, Korea Seung Ki Baek BK21 Physics Research Division and Department of Physics, Sungkyunkwan University, Suwon, 440-746, Korea Il Gu Yi, Hye Jin Park & Beom Jun Kim Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Contributions B.S.K. & B.J.K. designed research, performed research, wrote, reviewed and approved the manuscript. I.G.Y. & H.J.P. performed the numerical and statistical analysis of the data. Ethics declarations Competing interests The authors declare no competing financial interests. Rights and permissions This work is licensed under a Creative Commons Attribution-NonCommercial-ShareALike 3.0 Unported License. To view a copy of this license, visit Reprints and permissions About this article Cite this article Baek, S., Yi, I., Park, H. et al. Universal statistics of the knockout tournament. Sci Rep 3, 3198 (2013). Download citation Received: 19 August 2013 Accepted: 28 October 2013 Published: 12 November 2013 DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Provided by the Springer Nature SharedIt content-sharing initiative Advertisement Explore content About the journal Publish with us Search Quick links Scientific Reports (Sci Rep) ISSN 2045-2322 (online) nature.com sitemap About Nature Portfolio Discover content Publishing policies Author & Researcher services Libraries & institutions Advertising & partnerships Professional development Regional websites © 2025 Springer Nature Limited Sign up for the Nature Briefing newsletter — what matters in science, free to your inbox daily.
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Coordinate and Trigonometry Bashing Justin Stevens Introduction: Coordinate bashing and trigonometry bashing are two very nice methods to solve geometry problems if you don’t know how to synthetically solve them. These methods are more applicable to AMC 10/12, AIME, Mandelbrot, and other introductory/intermediate contests. For Olympiad problems, look at the Baycentric Coordinates in the Olympiad Article forum on artofproblemsolving.com. Coordinate Bashing To coordinate bash, you set points to specific coordinates, and typically have one key point being (0,0) (as it is easy to use). From here, you write all given correspondences between side lengths and angles to complete your coordinate system. For intersection lines: Find the equation for the lines and find the intersection point by finding the intersection of the equations. From here, you can use trigonometry to find desired angles, or the distance formula to find desired side lengths. For perpendicular lines: Use the fact that the slopes are negative reciprocals to find the slope of the line. If you know any other facts about the perpendicular line, you can use the y − y0 = m(x − x0) equation. For parallel lines: Use the fact that they have the same slope. For right triangles: If a right triangle has legs a, b and hypotenuse c, put the triangle on the cartesian system with points at (0,0), (0,a), (b, 0) (and √a2 + b2 = c or a2 + b2 = c2). For equilateral triangles: Righting the triangle in the form ( −a 2 , −√3∗a 6 ), ( a 2 , −√3∗a 6 ), (0 , √3∗a 3 ) is helpful. The circumcenter turns out to be (0 , 0) which is the same as the incenter and orthocenter in an equilateral triangle. For other triangles (finding circumcenter/orthocenter): Set up equations to find the circumcenter and orthocenter. For example, if you have a triangle with coordinates (0,0), (5,0), (7,1), you could find the circumcenter by finding the perpendicular line to (0,0) and (7,1) and to (0,0) and (5,0) and finding the intersection. You can also set up the same equations with altitudes knowing the slope and a point that goes through the line. 2Let’s start out with an easy example and work our way up to some harder problems. Example 1: In circle O, P O ⊥ OB , and P O equals the length of the diamter of circle O. Compute P A AB . (Source: 1999 ARML Individual Round) Solution: OBAP Write O as (0 , 0), B as (0 , −r), and P as (2 r, 0), where r is equal to the radius. Remark that line BP is y = 1 2 x − r,and we desire to find the intersection of this with y2 + x2 = r2. Substitute this value for y into the above equation to give us 1 4 x2 − xr + r2 + x2 = r2 =⇒ 5 4 x2 = xr or x = 4 5 r (since x can’t be 0). Plug this into the equation for y to give us y = − 3 5 r, so the coordinates of the intersection of BP and circle O (which is A) is ( 4 5 r, −3 5 r). Now, remark that AB = √ 16 25 r2 + 4 25 r2 = r 5 √20. Also, AP = √ 36 25 r2 + 9 25 r2 = r 5 √45. Therefore AP AB = r 5 √45 r 5 √20 which is the same thing as √ 45 20 = 3 2 . Example 2: In △ABC , ∠C is a right angle. Point M is the midpoint of AB , point N is the midpoint of AC , and point O is the midpoint of AM The perimeter of △ABC is 112 and ON = 12 .5. What is the area of M N CB ?(Source: Mathcounts) Solution: Create the following diagram: (4 b, 0) (0 , 4a)ABCO (b, 3a) M (2 b, 2a)(0 , 0) (0 , 2a) N Remark that we have ON = 12 .5 from the problem, and using the distance formula, we get √a2 + b2 = 12 .5 = 25 2 .3Also, we have the perimeter of △ABC being equal to 112, so hence we must have 4 a + 4 b + 4 √a2 + b2 = 112. This implies that 4 a + 4 b = 62 = ⇒ a + b = 31 2 .From the diagram, add line N M , and draw in M M ′ such that M M ′ ⊥ CB . (4 b, 0) (0 , 4a)ABCO (b, 3a) M (2 b, 2a)(0 , 0) (0 , 2a) N (2 b, 0) M′ Remark that [ N M CB ] = [ N M M ′C] + [ M M ′B] = 4 ab + 2 ab = 6 ab .We have a2 + b2 = ( 25 2 )2, so hence we get ( a + b)2 − 2ab = ( 25 2 )2 =⇒ ( 31 2 )2 − 2ab = ( 25 2 )2.This implies that 2 ab = ( (31 −25)(31+25) 4 which implies that 2 ab = ( 6∗56 4 ) = 84, so hence we have [N M CB ] = 6 ab = 3(84) = 252 . Example 3: Square AIM E has sides of length 10 units. Isosceles triangle GEM has base EM , and the area common to triangle GEM and square AIM E is 80 square units. Find the length of the altitude to EM in △GEM . (Source: 2008 AIME I) Solution: Let E = (0 , 0), M = (10 , 0), I = (10 , 10) and A = (0 , 10). Let G = (5 , M ), and we look at the following diagram: E MIAGH1 H2 Lemma: △AEH 1 ∼= △IM H 2 Proof: AE = IM from square AIM E , ∠EAI = ∠AIM = 90 ◦ and 4∠AEH 1 = 90 − ∠H1EM = 90 − ∠H2M E from isosceles △GEM . Therefore △AEH 1 ∼= △IM H 2 by AAS .Remark that line EG is the equation y = m 5 x, and hence to find the coordinate of point H1, we use the equation y = m 5 x and y = 10 to give us x = 50 m . Hence, the coordinate of H1 is ( 50 m , 10), and we have [AH 1E] = 1 2 ∗ 10 ∗ 50 m = [ IH 2M ]. Since [ ABCD ] − [AH 1E] − [IH 2M ] is equal to the the area common to triangle GEM and square AIM E , so hence we have 100 − 2 ∗ 1 2 ∗ 10 ∗ 50 m = 80 = ⇒ m = 25. We desire to find the distance from G to EM , which is equal to the distance from (5 , 25) to (5 , 0) or 25 . Example 4: ABC is an equilateral triangle with side length 1. Point D lies on AB , point E lies on AC , and points G and F lie on BC , such that DEF G is a square. What is the area of DEF G ? (Source: 2012 Stanford Math Tournament) Solution: B CAF GE D Put the points on the cartesian system, where B is at point (0 , 0), C at (1 , 0) and A at ( 1 2 , √3 2 ). From here, I drew the square such that we have DG ⊥ BC and EF ⊥ BC , so that we have D above G and E above F . From here, let G = ( a, 0), let F be ( b, 0), and we are going to have D being ( a, a √3) (because the equation of AB which is where D is on is y = √3x), and E being ( b, −√3 ∗ b + √3) (because the equation of AC where E is on is y = −√3x + √3). From DEF G being a square, we have to have DE = EF = DG , or b − a = −√3b + √3 = √3a.From this, we have −b + 1 = a =⇒ b = 1 − a. Substitute this into b − a = √3a to give 1 − a − a = √3a =⇒ a(2 + √3) = 1 = ⇒ a = 1 2+ √3 . From this, we know that one side length of the square is √3 2+ √3 = √3(2 −√3) 1 = 2 √3 − 3. We need to square this to get the area, which is (2 √3 − 3) 2 = 21 − 12 √3 . Example 5: An equilateral triangle ABC is inscribed in a circle. Points D and E are midpoints of AB and BC , respectively, and F is the point where ( DE )→ meets the circle. Find DE/EF . (Source: ARML) Solution: Set the circumcenter to be O, which is at the point (0 , 0). WLOG, we let each side length be 1, and we put the vertices from left to right as A, C, B to give us A = ( −1 2 , −√3 6 ), C = (0 , √3 3 ), B = ( 1 2 , − √3 6 ) (note that this is derived from originally putting the points at A = (0 , 0) , B = (1 , 0) and C = ( 1 2 , √3 2 ), from which we get O = ( 1 2 , √3 6 ) and then move this point to the origin.) Now remark that we desire to find DE EF , and we are going to use a combination coordinate bashing, and similar triangles to solve this problem. First off, remark that OA = OC = OB by definition of the circumcenter of △ABC . From this, we get the circumradius of the triangle is 1 3 , so hence the circle is defined by x2 + y2 = 1 3 . Also, remark that since D is the midpoint of AB , we have D = (0 , −√3 6 ), and similarly E = ( 1 4 , √3 12 ). 5Now, we find the slope of DE , and use that the y intercept is b = −√3 6 to find the equation of this line. The slope is √3 12 +√3 6 1 4 = √3 4 ∗ 4 = √3. So the equation of this line is y = √3x − √3 6 . Substitute this into the equation y2 + x2 = 1 3 to give 3( x2 − 1 3 x + 1 36 ) + x2 = 1 3 =⇒ 4x2 − x + 1 12 = 1 3 =⇒ 4x2 − x − 1 4 = 0 which gives 16 x2 − 4x − 1 = 0. From this, use the quadratic equation to find that x = 4±√16+64 32 = 4±4√5 32 = 1±√5 8 . Since we are looking for when ( DE )→ intersects the circle, we use the greater value of x, to give us the x coordinate being 1+ √5 8 of the point of intersection. Now is where we use the similar triangles. Look at segment DF , and drop E down to DB such that they meet at H and EH ⊥ DB . Also, drop F down to line DB such that DB ⊥ F G . Remark that we must have △F DG ∼ △ DEH by AA similarity, so hence we have DE DH = DF DG . Re-arrange this to give DE DF = DH DG . We know DH = 1 4 (since D = (0 , − √3 6 ) and H = ( 1 4 , − √3 6 ) and DG = 1+ √5 8 (same reasonign), so hence we get DE DF = 2 1+ √5 .Take the reciprocal of both sides to give DF DE = 1+ √5 2 , and DF = DE + EF , so we get DE +EF DE = 1+ √5 2 =⇒ EF DE = √5−1 2 . Now, we take the reciprocal of both sides once more to give DE EF = 2√5−1 = 2( √5+1) 4 = 1 + √5 2 .6Trigonometry Bashing Trigonometry bashing is done in problems that you are trying to find angles or side measures and you do not have a lot of information on the problem. Trigonometry bashing and coordinate bashing can be used together, which is a very useful tool, or alone. Trigonometry bashing has a few key steps which I will show. Law of sines/cosines A very useful tool in many problems is to use the law of sines and cosines. In a triangle with sides A, B, C with angles a, b, c opposite to their respective sides, sin( ∠a) A = sin( ∠b) B = sin( ∠c) C .For a triangle with the same conditions, c2 = a2 + b2 − 2ab cos( ∠c) Area formula for triangle The area of a triangle with the same conditions as before is 1 2 × ab sin( c) Sum of angles is 180 degrees This can be a useful tool if you have an isosceles triangle and you have side correspondences. Using the law of sines and the fact that sin(180 ◦ − θ) = sin( θ), you can find angle relationships with other angles. 7I will give two examples that show trigonometry bashing. Example 1: Let △ABC be a right triangle with right angle at C. Let D and E be points on AB with D betwen A and E such that CD and CE trisect ∠C. If DE BE = 8 15 , then tan B can be written as m√p n , where m and n are relatively prime positive integers, and p is a positive integer not divisible by the square of any prime. Find m + n + p. (Source: 2012 AIME I) Solution: See the following diagram created by David Altizio: BACED Note that ∠ACD = ∠DCE = ∠ECB = 30 ◦.From the law of sines: sin(30 ◦ ) 8x = sin( ∠CDE ) CE and sin(30 ◦ ) 15 x = sin( ∠DBC ) CE Hence 8 x sin( ∠CDE ) = 15 x sin( ∠DBC ) = sin(30 ◦)CE or 8 sin( ∠CDE ) = 15 sin( ∠DBC ). Note that ∠CDE + ∠DBC = 120 ◦ from △CDB having ∠DCB = 60 ◦. Hence, we get ∠CDE = 120 ◦ − ∠DBC .Substitute this in to give us 8 sin(120 ◦ − ∠DBC ) = 15 sin( ∠DBC ). This gives us 8 ∗ (sin(120 ◦) cos( ∠DBC ) − cos(120 ◦) sin( ∠DBC )) = 15 sin( ∠DBC ). This implies that 4√3 cos( ∠DBC ) + 4 sin( ∠DBC ) = 15 sin( ∠DBC ) which gives us: 4√3 cos( ∠DBC ) = 11 sin( ∠DBC ) = ⇒ sin( ∠DBC ) cos( ∠DBC ) = 4√3 11 which is the same as tan( ∠DBC ) or tan( B). This implies that we get an answer of 4 + 3 + 11 = 18 . Example 2: A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is √50 cm, the length of AB is 6 cm, and that of BC is 2 cm. The angle ABC is a right angle. Find the square of the distance (in centimeters) from B to the center of the circle. 8AB C Solution: Let O be the center of the circle, and draw line OC . By the Pythagorean theorem, we have AC = 2 √10. Note that in △AOC , we have ∠OAC = ∠ACO , which we let equal to θ and ∠AOC = β. Use the law of sines to give sin( θ) √50 = sin( β) 2√10 =⇒ sin( θ) = √5 sin( β) 2 . Also note that β + 2 θ = π which implies β = π − 2θ, so hence we get sin( θ) = √5 sin(2 θ) 2 =⇒ sin( θ) = √5 cos( θ) sin( θ). We divide this equation by sin( θ) since we can’t have sin( θ) = 0 (this would imply θ = 0), so hence we have cos( θ) = 1√5 = cos( ∠OAC )Also, remark that sin( ∠BAC ) = 2 2√10 = 1√10 by definition. We desire to find sin( ∠OAB ) = sin( ∠OAC − ∠BAC ) = sin( ∠OAC ) cos( ∠BAC ) − sin( ∠BAC ) cos( ∠OAC ). We know that sin( ∠OAC ) = 2√5 and cos( ∠BAC ) = 3√10 (from sin 2(θ) + cos 2(θ) = 1). Hence, we have sin( ∠OAB ) = 2√5 ∗ 3√10 − 1√10 ∗ 1√5 = 5√5∗√10 = 1√2 . From this, we have ∠OAB = 45 ◦.Now, finally, use the Law of cosines on △ABO to give us ( OB )2 = ( √50) 2 + 6 2 − 2 ∗ √50 ∗ 6 ∗ √2 2 = 86 − 60 = 26 .9
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https://www.geeksforgeeks.org/engineering-mathematics/application-of-partial-derivative-two-variable-maxima-and-minima/
Application of Partial Derivatives in Engineering Mathematics - GeeksforGeeks Skip to content Tutorials Python Java DSA ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps Software and Tools School Learning Practice Coding Problems Courses DSA / Placements ML & Data Science Development Cloud / DevOps Programming Languages All Courses Tracks Languages Python C C++ Java Advanced Java SQL JavaScript Interview Preparation GfG 160 GfG 360 System Design Core Subjects Interview Questions Interview Puzzles Aptitude and Reasoning Data Science Python Data Analytics Complete Data Science Dev Skills Full-Stack Web Dev DevOps Software Testing CyberSecurity Tools Computer Fundamentals AI Tools MS Excel & Google Sheets MS Word & Google Docs Maths Maths For Computer Science Engineering Mathematics Switch to Dark Mode Sign In Aptitude Engineering Mathematics Discrete Mathematics Operating System DBMS Computer Networks Digital Logic and Design C Programming Data Structures Algorithms Theory of Computation Compiler Design Computer Org and Architecture Sign In ▲ Open In App Application of Partial Derivatives in Engineering Mathematics Last Updated : 23 Jul, 2025 Comments Improve Suggest changes 1 Like Like Report Partial Derivatives can be used to find the maximum and minimum value (if they exist) of a two-variable function. We try to locate a stationary point with zero slope and then trace maximum and minimum values near it. The practical application of maxima/minima is to maximize profit for a given curve or minimize losses. Let f(x, y) be a real-valued function and let (pt, pt') be the interior points in the domain of f(x, y) then, pt, pt' is called a point of local maxima if there is an h > 0 such that f(pt, pt') ≥ f(x,y), for all x in (pt – h, pt' + h), x≠a The value f(pt, pt') is called the local maximum value of f(x,y). pt, pt' is called a point of local minima if there is an h < 0 such that f(pt, pt') ≥ f(x,y), for all x in (pt – h, pt' + h), x≠a The value f(pt, pt') is called the local minimum value of f(x,y). Partial Derivatives A partial derivative of a function of several variables is its derivative with respect to one of those variables, with all other variables held constant. For a function f(x,y), the partial derivative with respect to x, denoted as ∂f/∂x, measures the rate at which f changes as x changes, while y remains fixed. Partial derivatives are extensively used in engineering to model and solve problems involving multiple variables. These derivatives help in understanding how a system changes with respect to one variable while keeping others constant, providing essential insights into the behavior of physical systems. Algorithm to Find Maxima and Minima Find the values of x and y using fxx=0 and fyy=0 [NOTE: f xx and f yy are the partial double derivatives of the function with respect to x and y respectively.] The obtained result will be considered as stationary/turning points for the curve. Create 3 new variables r,t, and s. Find the values of r,t and s using r=fxx,t=fyy, s=fxy Perform the below operation : If(rt-s2)|(stationary pts)>0(Maxima/Minima) exists If (rt-s2)|(stationary pts)<0 (No Maxima/Minima)/(Saddle point) and now check for r below : Ifr >0 (Minima) If r <0 (Maxima) Applications of Partial Derivatives Here are the some applications of the partial derivatives in the engineering mathematics : Heat Transfer(Fourier's Law): In heat conduction, temperature varies with both time and space. Partial derivatives describe how temperature changes across a solid. Equation example: q = −k (∂T / ∂x) where q is heat flux, k is thermal conductivity, and ​∂T / ∂x is the temperature gradient. Fluid Dynamics(Navier-Stokes Equations): These partial differential equations describe the motion of fluid substances (liquids and gases). They involve velocity components, pressure, density, and viscosity. Application areas: weather forecasting, aerodynamics, and oceanography. Structural Analysis (Stress and Strain Analysis): Partial derivatives help determine how internal forces (stress) and deformations (strain) vary across a structure. Important for designing buildings, bridges, aircraft, and other load-bearing systems. Electromagnetics (Maxwell’s Equations): These fundamental equations describe how electric and magnetic fields evolve and interact. They rely on partial derivatives with respect to both space and time. Essential in: electrical engineering, telecommunications, antenna design, etc. Optimization Problems (Maximizing Efficiency / Minimizing Cost): Engineers use partial derivatives to find critical points of multivariable functions representing system performance. First-order derivatives locate critical points. Second-order partial derivatives help classify them as minima, maxima, or saddle points. Applications: manufacturing, energy systems, logistics, etc. Examples on the above Applications Example 1 : The function f(x,y)=x 2 y−3xy+2y+x has (a) No local extremum. (b) One local minimum but no local maximum. (c) One local maximum but no local minimum. (d) One local minimum and one local maximum. Solution: r = ∂2 f / ∂x 2 = 2y s = ∂2 f / ∂x ∂y = 2x−3 t = ∂2 f / ∂y2 = 0 Since, rt−s 2≤0, (if rt-s 2< 0 then we have no maxima or minima, if = 0 then we can't say anything). Maxima will exist when rt−s 2>0 and r<0. Minima will exist when rt−s 2>0 and r>0. As rt−s2is never greater than 0 so we have no local extremum. Answer: A Example 2 :Find the local minima of the function f(x , y) = 2x 2 + 2xy + 2y 2 - 6x f x(x,y) = 4x + 2y - 6=0 (1) f y(x,y) = 2x + 4y=0 (2) On solving (1) and (2) we get, x=2,y=-1 r = ∂2 f / ∂x 2 = 4 s = ∂2 f / ∂x∂y = 2 t = ∂2 f / ∂y 2=4 rt−s 2=12 As rt−s 2>0 and r>0. Answer : (2,-1) is the point of local minima. Example 3 :Find the maxima/minima of f(x , y) = x 2+y 2 + 6x +12 f x(x,y) = 2x+6=0 (1) f y(x,y) = 2y=0 (2) On solving (1) and (2) we get, x=-3,y=0 r=∂2 f/∂x 2=2 s=∂2 f/∂x∂y=0 t=∂2 f/∂y 2=2 As rt−s 2>0 and r>0. Answer : (-3,0) is the point of local minima Example 4 :Heat Conduction Problem:The temperature T(x,y) of a thin plate is given by T(x,y) = x 2 + 2xy + y 2. Find the rate of change of temperature with respect to x at the point (1,2). Solution: ∂T/∂x = 2x + 2y At (1,2): ∂T/∂x = 2(1) + 2(2) = 6 The rate of change of temperature with respect to x at (1,2) is 6 units/x. Answer : 6 units/x Example 5 : Fluid Dynamics Problem: The velocity field of a fluid is given by v(x,y) = 3x 2 y i + (x 3 + y 2) j. Find the acceleration in the x-direction at the point (2,1). Solution: ax = ∂vx/∂t + vx(∂vx/∂x) + vy(∂vx/∂y) ∂vx/∂x = 6xy ∂vx/∂y = 3x 2 At (2,1): ax = 0 + (12)(6) + (5)(6) = 102 units/s 2 Answer : 102 units/s 2 Example 6 :Optimization in Structural Engineering Problem :The deflection of a beam is given by w(x,t) = (L 3 / 48EI) (4x 3/L 3 - 3x/L) sin(πt/T), where L is the length, E is Young's modulus, I is the moment of inertia, and T is the period. Find the maximum deflection with respect to x. Solution: ∂w/∂x = (L 3 / 48EI) (4x 3/L 3- 3/L) sin(πt/T) Set ∂w/∂x = 0: 12x 2/L 3 - 3/L = 0 x 2 = L 2/4 x = L/2 The maximum deflection occurs at the midpoint of the beam. Answer : x = L/2 Example 7 : Electromagnetics Problem:The electric potential in a region is given by V(x,y,z) = 3x 2 y - 2yz 2 + 5xz. Find the electric field components at (1,1,1). Solution: Ex = -∂V/∂x = -6xy - 5z Ey = -∂V/∂y = -3x 2 + 2z 2 Ez = -∂V/∂z = 4yz - 5x At (1,1,1): Ex = -11, Ey = -1, Ez = -1 The electric field at (1,1,1) is E = -11i - j - k. Answer : E = -11i - j - k. Example 8 : Thermodynamics Problem :The pressure P of an ideal gas is given by P(V,T) = nRT/V, where n is the number of moles, R is the gas constant, V is volume, and T is temperature. Find (∂P/∂V)T and (∂P/∂T)V. Solution: (∂P/∂V)T = -nRT/V 2 (∂P/∂T)V = nR/V Example 9 : Control Systems Problem:A transfer function G(s) = K / (s 2 + 2ζωns + ωn 2) represents a second-order system. Find ∂G/∂K and ∂G/∂ζ. Solution: ∂G/∂K = 1 / (s 2+ 2ζωns + ωn 2) ∂G/∂ζ = (-2Kωns) / (s 2 + 2ζωns + ωn 2)2 Example 10 : Elasticity Problem:The strain energy density of a material is given by U(ε1,ε2) = (E/2) (ε1 2 + ε2 2 + 2νε1ε2), where E is Young's modulus, ν is Poisson's ratio, and ε1, ε2 are principal strains. Find the stress σ1. Solution: σ1 = ∂U/∂ε1 = E(ε1 + νε2) Practice Questions The displacement of a particle is given by s(t) = 3t 2 - 2t + 1. Find the velocity and acceleration at t = 2. The pressure in a fluid is given by P(x,y,z) = 5x 2 y + 3yz - 2z 2. Calculate the pressure gradient at the point (1,2,3). A company's profit function is P(x,y) = 100x + 80y - 2x 2 - xy - y 2, where x and y are the quantities of two products. Find the values of x and y that maximize profit. The temperature distribution in a plate is T(x,y) = 100 - 2x 2 - 3y 2. Find the direction of maximum temperature decrease at the point (2,1). The magnetic field in a region is given by B(x,y,z) = 2xyi + (x 2 - z 2)j + 3yzk. Calculate curl B at (1,1,1). A chemical reaction rate is given by r(T,C) = k0 exp(-E/RT) C n, where T is temperature and C is concentration. Find ∂r/∂T and ∂r/∂C. The deflection of a membrane is w(x,y) = A sin(πx/L) sin(πy/W), where L and W are the membrane dimensions. Find the points of maximum deflection. In a heat exchanger, the temperature difference is ΔT(x,t) = ΔT0 exp(-Ux/mc) (1 - exp(-t/τ)), where U, m, c, and τ are constants. Find ∂(ΔT)/∂x and ∂(ΔT)/∂t. The stress-strain relationship for a material is σ = E (ε - α ΔT), where E is Young's modulus, α is the thermal expansion coefficient, and ΔT is the temperature change. Find ∂σ/∂ε and ∂σ/∂T. A control system has a transfer function G(s) = K / (Ts + 1)2. Find ∂G/∂K and ∂G/∂T. 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https://math.stackexchange.com/questions/945356/combinatorial-argument-for-the-sum-of-the-first-n-integers
Skip to main content Combinatorial argument for the sum of the first n integers. Ask Question Asked Modified 10 years, 11 months ago Viewed 3k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Can someone give a combinatorial argument (at least for (n+12)) for why (n+12)=(n2+n)/2? combinatorics summation combinatorial-proofs Share CC BY-SA 3.0 Follow this question to receive notifications edited Sep 25, 2014 at 9:34 Martin Sleziak 56.2k2020 gold badges210210 silver badges391391 bronze badges asked Sep 25, 2014 at 6:45 user170141user170141 32333 silver badges99 bronze badges 6 1 Now I'm notationally baffled. I know that nCr is sometimes used fo (nr), but what is nCr(n+1,2) supposed to mean? – Hagen von Eitzen Commented Sep 25, 2014 at 7:07 I was just unsure about how to enter a binomial coefficient here... I just mean n+1 choose 2. Sorry for the confusion. – user170141 Commented Sep 25, 2014 at 7:10 2 I think that several answers from older posts about this sum qualify as combinatorial proofs. For example N.S.'s answer, Qiaochu's answer, yoyo's answer from (probably) the most popular post asking for proof of this identity. – Martin Sleziak Commented Sep 25, 2014 at 9:37 Just to clarify: You are asking for a combinatorial argument for 1+2+⋯+n=(n+12) (as your title indicates), right? Since when I read the body of your question, it seems as if you were asking why (n+12)=n2+n2. – Martin Sleziak Commented Sep 25, 2014 at 9:41 @MartinSleziak: Your last comment is spot on! I just thought about the same confusion. It seems to me we are considering two seperate questions that maybe the OP thinks make up a perfect entity. – String Commented Sep 25, 2014 at 9:45 | Show 1 more comment 5 Answers 5 Reset to default This answer is useful 4 Save this answer. Show activity on this post. Here is a standard way to view this identity: There are n+1 people in a room. Everyone shakes hands with everyone else (one handshake per pair of people; nobody shakes his or her own hand). Let's count how many handshakes occurred: Method 1: For each pair of people there is one handshake. So there are (n+12) handshakes. Method 2: Each of the n+1 people reports having been involved in n handshakes. This gives (n+1)⋅n reported handshakes. However this counts each handshake twice (once by each of its participants). So there are (n+1)⋅n2 handshakes. Method 3: Line up the people in a row. First person shakes everyone else's hand: n shakes. The next person shakes hands with everyone except the first person: n−1 new shakes. The next person shakes hands with everyone except the first two people: n−2 new shakes. And so on. This gives a total of n+(n−1)+(n−2)+⋯+1 handshakes. Comparing the three ways of viewing this situation, we have (n+12)=(n+1)⋅n2=1+2+3+⋯+n Share CC BY-SA 3.0 Follow this answer to receive notifications answered Sep 25, 2014 at 10:08 paw88789paw88789 41.9k33 gold badges3838 silver badges7878 bronze badges Add a comment | This answer is useful 1 Save this answer. Show activity on this post. let you have n+1 balls(with different color) you have to select 2 balls and put it in a box selecting first ball has n+1 ways and second ball has n ways total (n+1)⋅n ways but as you are putting ball in a box it's order is not important this means your are counting twice this leads to (n+1n)=(n+1)n2=n2+n2. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Sep 25, 2014 at 9:46 String 18.8k55 gold badges4545 silver badges8484 bronze badges answered Sep 25, 2014 at 8:19 HarishHarish 14922 silver badges88 bronze badges Add a comment | This answer is useful 0 Save this answer. Show activity on this post. For any integer n>1, there are n−1 ways to write n as a sum n=a+b of two strictly positive integers a,b>0 (ignoring order). For example, 4=3+1=2+2=1+3 makes three ways. Let us now count the number of ways S3(n) to write n>2 as a sum n=a+b+c of three strictly positive integers a,b,c>0. To do so, let us pick m such that 1≤m≤n−1. We can write k=n−m as a sum of two strictly positive integers k=b+c in k−1=n−m−1 ways so we conclude that we can write n=m+b+c (m,b,c>0) in the following number of ways: S3(n)=∑m=1n−1(n−m−1)=∑k=1n−1(k−1)=∑k=1n−2k. Another way to count S3(n) is to consider n objects labelled with 1,2,…n, pick two distinct numbers p and q, 1≤p<q≤n−1 and make three groups: one group with the objects labelled from 1 to p, one group with the objects labelled from p+1 to q and the third group with the remaining objects. There are by definition (n−12)=(n−1)(n−2)2 ways to do this. We have now proven that S3(n+1)=(n2)=n(n−1)2=∑k=1n−1k. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Sep 25, 2014 at 8:10 Tom-TomTom-Tom 7,04111 gold badge2020 silver badges4747 bronze badges Add a comment | This answer is useful 0 Save this answer. Show activity on this post. To see combinatorially how the sum of the first n integers match the number of ways to pick 2 elements, (a,b), from the list 1,2,...,(n+1), you could as follows: Fix one element, a, as you first choice Pick the next element b to be strictly greater than a If a=1 you have n choices for b. If a=(n+1) you have zero choices for b since then it cannot be greater than a. Thus the number of choices for b while a runs from 1 through (n+1) can be summarized as n+...+2+1+0 which happens to be the sum of the first n integers. As to why this should equal (n2+n)/2 I think Suraj posted a perfectly simple and clear explanation for that. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Apr 13, 2017 at 12:20 CommunityBot 1 answered Sep 25, 2014 at 9:43 StringString 18.8k55 gold badges4545 silver badges8484 bronze badges Add a comment | This answer is useful 0 Save this answer. Show activity on this post. Consider putting n-1 different balls in 3 bins. The stars and bars method would give the answer (n−1)+3−1Cn−1=n+1C2 as the answer. The other way we can distribute the balls is as follows: When we put all n−1 balls into one bin, the other two bins have total 0 objects in them. So for this case we have only 1 way to distribute the balls. When we put n−2 balls into one bin, we have one ball to put in the other two. There are two ways to distribute it among the other two bins (labelled as (number of balls in the first of the other two, number of balls in the other)) (0,1),(1,0). And so on, so when we put n−k objects into one bin, there are k+1 ways to distribute the rest of the balls into the other two bins, (k,0),(k−1,1),(k−2,2),...,(1,k−1),(0,k). Adding it all up we get: 1+2+⋯+n=n(n+1)2=(n+12) as required. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Sep 25, 2014 at 9:44 answered Sep 25, 2014 at 7:12 James HarrisonJames Harrison 1,09211 gold badge88 silver badges1616 bronze badges Add a comment | You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics summation combinatorial-proofs See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... Community help needed to clean up goo.gl links (by August 25) Linked 141 Proof that 1+2+3+4+⋯+n=n×(n+1)2 27 Is there a combinatorial interpretation of the triangular numbers? 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7473
https://thetandemramble.com/travel-tips/bicycle-touring/advantages-and-disadvantages-of-a-tandem-bicycle/
The Tandem Ramble Benjamin Nerding Advantages and Disadvantages of a Tandem Bicycle After traveling many years by tandem bike we wanted to share some tips with you as an upcoming tandem cyclist. If you are unsure of whether to take a tandem bike or two single bikes for your next trip with your partner, this is the right place to be. We have summarized 12 pros and 9 cons to make your choice easier. The decision will follow you for a long time after, so choose wisely. Tandem Bicycle or Two bicycles. What´s better? Here are the Advantages and Disadvantages of a Tandem Bicycle: Advantages of a Tandem Bicycle: 2 wheels, 2 breaks, 2 seats Basically, every mechanical issue exists only once as you carrying two people with only one bike. The practicality is that everything that can go bad on a bike trip is at least to some extent minimized. Show your glory and use the attention Your bike will attract a lot of attention. In our point of view, people tend to respect you more like a proper vehicle to share the road with. Of course, this heavily depends on your travel destination but generally, you will more likely receive help. TEAM – Together everyone achieves more Since you are riding all the time together, you achieve something as a couple. This accomplishment not only feels great but also creates bonds between the two riders. In the plain, feel less pain Riding in the plain can be an advantage compared with a normal bicycle as you are more aerodynamic than 2 bikes and less heavy. One disconnects doubles the effects It saves energy of the co-pilot when riding the bike for a longer period. The stoker only has to peddle and sit in the seat. The pilot has to be concentrated. So one person less to do the steering job. ‘You keep steering, I do the rest’ All tasks that are done while biking is divided in two. As the pilot steers the bike the stoker is free to take pictures, navigate or supply food. No time wasted You can do your breaks for anything simultaneously and therefore don’t lose time. No need to wait, he/she is already here You are always together, so there is no need to wait. So nobody needs to get annoyed. Also, depending on the country you travel in, you can protect one another. Take one for the team The stronger member can balance out the weaker member and everyone gets his exercise equally after all. 4 eyes see more than 2 Two people on one bike mean having 4 eyes to look out for the traffic instead of 2. So that taxi cutting you off might be spotted just a second earlier. Talk the talk In long hours of silence, you might want to share something which is on your mind with your travel buddy. Theft secure A tandem bicycle is much less likely to be stolen as it attracts a lot of attention. It is also very uncommon in most countries. Advantages and Disadvantages of a Tandem Bicycle Disadvantages of a Tandem: Big and clunky The tandem bike is anything but gracious riding. The bike is longer and heavier which makes it unhandy. Keeping the balance, starting at the same time and riding is a lot more challenging with an extra person on your back seat. ‘The back isn’t peddling’ A sentence you will have heard in many languages on many occasions from many many people. Your constantly receiving full attention from pretty much anyone that sees you. This also includes the honking of cars in many countries. Carriage requirements – single seat Many airlines include this into their special baggage requirements which makes the tandem bicycle a tricky travel companion. Our experience in South America involved many times calling different airline call centers and being on hold (sometimes for hours). The hills are killing Much more difficult to ride a tandem bicycle if hilly or in mountain areas or against wind and downhill, off-road. Share the pain If one of you falls, the other one will too. Maneuvering the tandem bike through areas can be tricky and falling with a tandem bike might leave both riders injured, let’s hope someone stops and helps. ‘I don’t feel like it today’ If one person does not want to ride, the tandem becomes a burden. If you travel with two bikes, it is much easier to do stuff by yourself. Traveling with a partner can be tiring, you might want some time alone. More load = more trouble While bike packing with a tandem bicycle you certainly need hydraulic breaks that are much more likely to be tested to its limits compared to two single bikes. Money is the issue? Tandem bikes are more expensive than 2 normal bikes and can only be used by two people. You need an extra-long brake and gear cables which make reparations much harder. Should I stay or should I go? In case you have a major equipment problem during your travels. It is very difficult to hitchhike with a tandem bicycle or it might even be challenging to take a bus. One stays behind and the other one leaves? We recommend an extremely good Tandem manufacturer! Tandems are much more injury-prone than normal bicycles. You need a good frame and good material to keep your bicycle healthy for a long time during your travels. Cycling! Now, enjoy your time together and discuss the options with your partner. Taking a decision together is the most important part of traveling with other people. You can certainly have fun with a tandem bicycle and 2 bicycles. The decision depends very much on your individual situation, the length of your travels, and your fitness level. This was our article about the Advantages and Disadvantages of a Tandem Bicycle. Read everything about 5 simple Tips to Prevent Cycling Pains During And After A Bicycle Ride. Follow us on Instagram. 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7474
http://www.witno.com/philadelphia/notes/trial.html
| | | | --- | | At first you don't succeed. Try, try again. « Witno.com Logic Quote | | | | AMIN WITNO WEB PHILADELPHIA UNIVERSITY | | | | | | | | | | | --- --- --- --- | | | | | HOME | | COURSES | | NOTES | | EXAMS | | RÉSUMÉ | | TRIAL DIVISION TO TWELVE DIGITS What is Trial Division? Trial division is the name of an algorithm which attempts at finding a prime factor of a given positive integer N. The trial is done by repeatedly dividing N by the prime numbers 2, 3, 5, 7, 11, etc., until a factor is found; otherwise N itself is prime. Note that given a composite N which factors into two or more primes, at most one of the prime factors can be larger than √N while the rest is each bounded by this square root. Therefore, it is not necessary to continue the algorithm once we reach the last prime number p ≤ √N. This Implementation Trial division is very easy to implement in any programming language. In this JavaScript version we store an array of all prime numbers up to 1,000,000. There are exactly 78,498 such primes, the largest of which is p = 999,983. Thus we are set to take on any integer N up to one trillion, i.e., twelve decimal digits, since √1012 = 1,000,000. Furthermore, once we find a prime factor p, we iterate the trial division with the integer N/p in place of N. In this way, we end up with the complete factorization of N into prime numbers. As already mentioned, if not one factor is found, then the number N proves to be a prime. JavaScript integers can assume up to 53 bits, where 253 = 9,007,199,254,740,992. In the implementation, we let any N in this range be tested until p = 999,983. However, if the unfactored part remains larger than a trillion by the end of the iterations, then we will be content with an "incomplete" factorization. Feel free to save or modify a copy of this file for your own use. Other Factoring Algorithms Despite its simple implementation, trial divison is super slow when dealing with large numbers. To tackle a 53-bit integer, for instance, our prime database would swell 70 to 80 times larger than its current size. For a more advanced and more clever factoring algorithms, try to google the so-called Pollard rho method and p−1 method, or the quadratic sieve method. None of these guarantees a quick result, if successful at all, but with some theoretical help, one can learn the best factoring approach with respect to a given integer range. There are situations in which one must know simply whether N is prime or composite, without needing the factors of N if composite. This calls for a primality testing algorithm. Generally speaking, testing primes is not as hard as factoring composites, nor as slow. See our testing primes page for an example. Copyright © 2011–2023 Amin Witno --- This page belongs to the personal folder of Amin Witno and does not necessarily represent the philosophy and values of Philadelphia University or the Department of Basic Sciences in particular. | | | |
7475
https://www.openmiddle.com/multiplying-two-digit-numbers-elementary/
Multiplying Two-Digit Numbers (Elementary) | Open Middle® Home Kinder Counting & Cardinality Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry 1st Gr Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry 2nd Gr Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry 3rd Gr Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions Measurement & Data Geometry 4th Gr Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions Measurement & Data Geometry 5th Gr Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions Measurement & Data Geometry 6th Gr Ratios & Proportional Relationships The Number System Expressions & Equations Geometry Statistics & Probability 7th Gr Ratios & Proportional Relationships The Number System Expressions & Equations Geometry Statistics & Probability 8th Gr The Number System Expressions & Equations Functions Geometry Statistics & Probability High School Number & Quantity The Real Number System Quantities The Complex Number System Vector and Matrix Quantities Algebra Seeing Structure in Expressions Arithmetic w/ Polynomials & Rational Expressions Creating Equations Reasoning with Equations and Inequalities Functions Interpreting Functions Building Functions Linear, Quadratic, and Exponential Models Trigonometric Functions Geometry Congruence Similarity, Right Triangles, and Trigonometry Circles Expressing Geometric Properties with Equations Geometric Measurement and Dimension Statistics & Probability Interpreting Categorical and Quantitative Data Making Inferences and Justifying Conclusions Conditional Probability & the Rules of Probability Using Probability to Make Decisions Calculus Computer Science Physics About Articles What’s Open Middle? Open Middle Team Submit English English Español Français Take The Online Workshop Open Middle® Take The Online Workshop Home Kinder Counting & Cardinality Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry 1st Gr Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry 2nd Gr Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry 3rd Gr Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions Measurement & Data Geometry 4th Gr Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions Measurement & Data Geometry 5th Gr Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions Measurement & Data Geometry 6th Gr Ratios & Proportional Relationships The Number System Expressions & Equations Geometry Statistics & Probability 7th Gr Ratios & Proportional Relationships The Number System Expressions & Equations Geometry Statistics & Probability 8th Gr The Number System Expressions & Equations Functions Geometry Statistics & Probability High School Number & Quantity The Real Number System Quantities The Complex Number System Vector and Matrix Quantities Algebra Seeing Structure in Expressions Arithmetic w/ Polynomials & Rational Expressions Creating Equations Reasoning with Equations and Inequalities Functions Interpreting Functions Building Functions Linear, Quadratic, and Exponential Models Trigonometric Functions Geometry Congruence Similarity, Right Triangles, and Trigonometry Circles Expressing Geometric Properties with Equations Geometric Measurement and Dimension Statistics & Probability Interpreting Categorical and Quantitative Data Making Inferences and Justifying Conclusions Conditional Probability & the Rules of Probability Using Probability to Make Decisions Calculus Computer Science Physics About Articles What’s Open Middle? Open Middle Team Submit English English Español Français Home>Grade 4>Multiplying Two-Digit Numbers (Elementary) Multiplying Two-Digit Numbers (Elementary) Directions: Using the digits 1 to 9 at most one time each, fill in the boxes to make the smallest (or largest) product. Hint What does the number on the left represent? What does the number on the right represent? Answer 96· 87 is one answer for the largest product. 13· 24 is the smallest product. Source: Robert Kaplinsky Embed this problem Print Embed This Problem X [x] Show Title [x] Show Author [x] Make Problem Clickable [x] Show Hint [x] Show Answer Embed Code: Copy Close Tags 4.NBT.55.NBT.5DOK 3: Strategic ThinkingRobert Kaplinsky Previous Multiplying Decimals (Elementary) Next Subtracting Two-Digit Numbers (Elementary) Check Also Add and Subtract Mixed Numbers Directions: Using the digits 1-9 at most one time each, create an equation using addition … 22 comments alexDecember 13, 2016 at 11:47 am 98×76 Reply Robert KaplinskyFebruary 2, 2017 at 11:04 am That’s pretty close Alex! Can you make it any larger? Reply lukeOctober 18, 2017 at 8:05 am 96 x 87 and 13 x 24 Reply BrigApril 23, 2018 at 9:13 am 96 x 87 and 13 x 24 Reply Rusty AndersonFebruary 6, 2017 at 6:46 am Here is a Desmos visual that might support the task. Reply Robert KaplinskyMay 24, 2017 at 10:20 am Nice Rusty! Too bad there isn’t a built in way to restrict duplicate numbers. Two thoughts: – I wonder if there is a difference between a horizontal and vertical representation? – Is there value in restricting a and b to 12 to 98? (11 or less and 99 and 100 are not possible) Reply Raiden From 3rd GradeFebruary 27, 2018 at 6:47 am I Did 12×5 I Dint Try For Largest Or Smallest My Whole Answer Was This: 12+12+12+12+12= 60 24 24 12=60 24+24+12=60 48+12=60 Reply 4. TristainMarch 30, 2020 at 1:44 pm 24×13=212 87×96=1152 Reply 5. KeyannaApril 1, 2020 at 1:03 pm 1234=38 and 1025=25 Reply 6. GiannaApril 1, 2020 at 2:11 pm 96×87 is the biggest. Reply 7. SerenaApril 2, 2020 at 4:17 pm 14×23=322 Reply 8. michael TROTTERApril 7, 2020 at 7:38 pm 18 times 28=2204 Reply 9. ledger leeApril 8, 2020 at 7:36 am 3234=1,088 Reply 10. evelynApril 9, 2020 at 10:52 am 96 x 87=8,282 Reply 11. MatthewSeptember 22, 2020 at 5:59 am 97×86 biggest 13×24 smallest Reply 12. shawnSeptember 22, 2020 at 10:32 am 98 x 76 largest 1324 Reply 13. AliceMarch 16, 2021 at 7:59 am 96×87 Reply 14. LandonOctober 13, 2022 at 9:42 am 96×87 Reply 15. also LandonOctober 14, 2022 at 9:40 am smallest is 13 x 24 Reply 16. willlllllOctober 18, 2024 at 11:44 am 96 x 81 Reply 17. willlllllOctober 18, 2024 at 11:45 am sorry 87 x 96 Reply 18. willlllllOctober 18, 2024 at 11:46 am smallest 13 x 24 Reply Leave a Reply Cancel reply Your email address will not be published.Required fields are marked Comment Name Email Website [x] Save my name, email, and website in this browser for the next time I comment. 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7476
https://math.stackexchange.com/questions/1856556/finding-ratio-of-cevian-lines
Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Finding ratio of cevian lines Ask Question Asked Modified 4 years, 11 months ago Viewed 2k times 1 $\begingroup$ I am preparing for an exam and doing some pratice problems. So I'm having a difficult time with this problem. At first I thought the ratio was 2:1 and then I also thought I would be able to use the angle bisector theorem but then I realized that cevians could be altitudes, medians or angle bisectors. So theorems are not helpful with this problem. Basically, I asked my teacher for help and he told me that the answer is EF:FC is $10:7$ and DF:FA is $3:14$ I've been trying to find out how he got these ratios but i'm not getting anywhere. Any ideas? geometry projective-geometry Share edited Jul 12, 2016 at 1:17 Ted Shifrin 127k77 gold badges113113 silver badges174174 bronze badges asked Jul 12, 2016 at 0:56 mikamika 85788 silver badges2222 bronze badges $\endgroup$ 0 Add a comment | 4 Answers 4 Reset to default 2 $\begingroup$ After joining BF, label the areas as shown. For example, p = area of triangle BEF = [BEF]. The rule:- €œ$\triangle$s having the same altitude, ratio of their areas is proportional to the ratio of their bases.€ For example, $\dfrac {p}{z} = \dfrac {3}{4}$. This means $p = \dfrac {3}{4}z$. Do the same to $\dfrac {q}{x} = €¦$ to get $q = ? x$. Therefore, $[BEFD] = p + q = \dfrac {3}{4}z + ?x$. Apply the same rule to get $\dfrac {[ABD]}{[ADC]} = \dfrac {5}{2}$. Combining all the results, you should get $\dfrac {z}{y} = \dfrac {10}{7}$ (with x disappears miraculously). Apply the converse of the rule to get $\dfrac {EF}{FC} = €¦ $. Share answered Jul 12, 2016 at 3:35 MickMick 17.5k44 gold badges3232 silver badges5656 bronze badges $\endgroup$ Add a comment | 2 $\begingroup$ Menelaus' theorem for $\triangle BCE$ and transversal $DA$ gives $ \frac{EF}{FC}\; \frac{CD}{DB}\; \frac{BA}{AE} = 1 = \frac{EF}{FC}\; \frac{2}{5}\; \frac{3+4}{4}$ thus $\frac{EF}{FC} = \frac{10}{7}$. Permute $5 \leftrightarrows 3, 2 \leftrightarrows 4$ to get $\frac{DF}{FA} = \frac{3}{14}$. Share answered Jul 12, 2016 at 5:08 dxivdxiv 78k66 gold badges6969 silver badges127127 bronze badges $\endgroup$ Add a comment | 1 $\begingroup$ A general method to solving these kinds of things is through the Ratio Lemma, which is often applied in high school olympiad geometry. It states that if $AD$ is a cevian in $\triangle ABC$, then $\dfrac{BD}{DC}=\dfrac{AB}{AC}\cdot\dfrac{\sin\angle BAD}{\sin\angle CAD}$. The proof of this is quite simple; just apply the sine law to triangles $ABD$ and $CAD$. So for this problem, the ratios $\sin\angle BAD:\sin\angle CAD$ is clearly equal to $\sin\angle EAF:\sin\angle CAF$. So $EF:FC$ is equal to $(BD/DC)(AE/AB)=10/7$. $DF:FA$ is calculated similarly. The Ratio Lemma can also be used to prove many basic facts in projective geometry; e.g. the invariance of the cross-ratio when projected through a point onto a line. Share answered Jul 12, 2016 at 2:06 CyclicduckCyclicduck 90155 silver badges1515 bronze badges $\endgroup$ Add a comment | 0 $\begingroup$ Use mass points. We assign B a mass of 4. So A has a mass of 3 and C has a mass of 10. E has a mass of 7 and D has a mass of 14. So we have $$ \frac{AF}{FD}= \frac{14}{3}$$ and $$\frac{EF}{FC}= \frac{10}{7}$$. Share answered Oct 29, 2020 at 7:48 user843256user843256 1 $\endgroup$ 1 $\begingroup$ You need to include more detail into what "mass points" are. $\endgroup$ Toby Mak – Toby Mak 2020-10-29 11:19:49 +00:00 Commented Oct 29, 2020 at 11:19 Add a comment | You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry projective-geometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 6 A problem on tangent and secant lines Related Constructing a triangle given three concurrent cevians? 2 Prove that the center of a circle within a constructed triangles lies on the angle bisector 1 Geometry problem with bisectors 0 Kiselev's geometry Problem 67: In an isosceles triangle, two medians/bisectors/altitudes are congruent 0 Intersecting triangles: Find the length of a segment given one side and two ratios 0 $ABC$ exists so $\overline{AX}$, $\overline{BC}$, and $\overline{CZ}$ are concurrent. $\overline{AB} = 6$, $\overline{AC} = 3$, $\overline {BX}$ is? 0 Proving that internal line segments are concurrent by Ceva's theorem for an incircle Prove a collinearity equivalence (Euclidean geometry) 3 Orthocenter: The "Bad Boy" of Distinguished Points in a Triangle Hot Network Questions Proving a certain Cantor cube is a complete metric space (by definition) - proof verification Lingering odor presumably from bad chicken Storing a session token in localstorage Bypassing C64's PETSCII to screen code mapping How to locate a leak in an irrigation system? 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Is it ok to place components "inside" the PCB в ответе meaning in context How to sample curves more densely (by arc-length) when their trajectory is more volatile, and less so when the trajectory is more constant Find non-trivial improvement after submitting Calculate center of object and move it to the origin and center it using geometry nodes more hot questions Question feed
7477
https://files.eric.ed.gov/fulltext/EJ744025.pdf
T he 1988 Extension 1 (3 unit) Mathematics New South Wales Higher School Certificate examination contained the following innovative ques-tion: Question 6 b) Suppose . i) Use the binomial theorem to write an expression for tk, 0 ≤k ≤25. ii) Show that . iii) Hence or otherwise find the largest coefficient tk. You may leave your answer in the form . Similar questions have made regular subsequent appearances in trial examinations around NSW and many texts now devote whole chapters to the subtle and somewhat laborious process of establishing the largest coefficient in a binomial expansion. This article will produce a closed form solution to all questions of this type. Along the way, we will point out the limitations of the above approach and investigate the intriguing manner in which the great-est coefficient moves about. The content of the article is accessible to advanced mathematics students in the final year of high school and is of particular value to Extension 1 and 2 students in the NSW Higher School Certificate. I have included a number of discussion points and extension problems to facilitate possible classroom discussions. We begin our investigation by considering the simpler expansion (2x + 1)6 = 1 + 12x + 60x2 + 160x3 + 240x4 +192x5 +64x6 with t0 = 1, t1 = 12, t2 = 60, t3 = 160, t4 = 240, t5 = 192 and t6 = 64. 53 Australian Senior Mathematics Journal 20 (1) Largest coefficients in binomial expansions Milan Pahor The University of New South Wales pahor@maths.unsw.edu.au The critical feature to note is that the coefficients rise and then fall in a controlled manner. That is 1 ≤12 ≤60 ≤160 ≤240 ≥192 ≥64. The largest coefficient is clear with the coefficients first rising to and then falling from 240. This behaviour is in fact typical of certain binomial expan-sions and it is a property we exploit to attack larger questions where a direct expansion is impractical. Comparing the ratio of each coefficient to its predecessor we have If the largest k for which is k = m then the largest coefficient is tm+1. In the above expansion, the largest k for which is k = 3 and thus the largest coefficient is t4 = 240. The general technique for establishing the maximal coefficient is built on this simple analysis of the ratio of successive binomial coefficients . [Discussion question: When attempting to maximise quantities we usually appeal to the calculus: why is it inappropriate to do so here?] We will need a curious but very handy little function over the real line called the floor function. Definition: floor(x) is the greatest integer less than or equal to x. Thus: floor(3.2) = 3, floor(–5.2) = –6, floor(π) = 3 and floor(2) = 2. The graph of the floor function is: Figure 1 [Discussion Question: What is the domain and range of the floor function? Describe the domain, range and graph of the derivative of floor(x).] We are now in a position to prove our first formula as a lemma. 54 Australian Senior Mathematics Journal 20 (1) Pahor Lemma Suppose that d is a positive real number and that n is a positive integer. Then the largest coefficient in the binomial expansion of (dx + 1)n is where Proof Using the Binomial theorem we have where Now So 55 Largest coefficients in binomial expansions Australian Senior Mathematics Journal 20 (1) which is true if (noting that both k + 1 and d + 1 are positive). Thus only for k = 0, 1… . Therefore the maximal coefficient is To complete the proof, note that and implying α is one of the integers 0, 1, 2… n. Thus is well defined. QED. Applying the above result to the expansion (2x + 1)6 = 1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6 we have d = 2 and n = 6. Thus The largest coefficient is then as expected! Fixing n and varying d It is fascinating to plot α versus d for a fixed n, say n = 6. The sketch of is shown in Figure 2. If we fix n and let d vary from 0 to n, the greatest coefficient in the expansion will elegantly march from the constant term x0 up to the final term xn. 56 Australian Senior Mathematics Journal 20 (1) Pahor Figure 2 We have seen above that 0 ≤α ≤n. Furthermore and [Discussion question: Show that for fixed n, is an increasing function of d.] Sampling a few expansions for n = 6 we have: (0.2x + 1)6 = 1 + 1.2x + 0.6x2 + 0.16x3 + 0.024x4 + 0.00192x5 + 0.000064x6 (x + 1)6 = 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6 (2x + 1)6 = 1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6 (5x + 1)6 = 1 + 30x +375x2 + 2500x3 + 9375x4 + 18750x5 + 15625x6 The discontinuities in Figure 2 are of course the critical values of d where the largest coefficient in the expansion of (dx + 1)6 leaps to its new home. When does this happen? We require . Thus: 57 Largest coefficients in binomial expansions Australian Senior Mathematics Journal 20 (1) It follows that the jumps occur at . For n = 6, the critical values are Very cute and beautifully balanced! These are the six values of d where the greatest coefficient in the expansion of (dx + 1)6 latches on to a new power of x. But how is the jump achieved? We have a few philosophical problems! The first is that the whole of binomial theory is characterised by symmetry and balance. What could possible justify one power of x over another at a critical value of d ? Furthermore for a particular power, say x4 of x, its coefficient is a continuous function of d. How can these continuous evolutions lead to discrete jumps? Well, let us take a look at the transition from (x + 1)6 = 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6 to (2x + 1)6 = 1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6. Somewhere between d = 1 and d = 2, the greatest coefficient has found its way from x3 to x4. From the above analysis this happens when d = . The expansion at that point is: The largest coefficient in the expansion slides along the expansion as d increases, replicating itself at the critical points where it needs to leap to the next power of x. A truly remarkable method of transport! Before presenting our last result let us consider the role that the restric-tion that d be positive plays in this analysis. Consider the expansion: (–3x +1)6 = 1 – 18x + 135x2 – 540x3 + 1215x4 – 1458x5 + 729x6. Observe that we no longer have the coefficients rising uniformly to a maximum and then falling! That is, the fundamental property driving our technique no longer holds. Note also that simply ignoring the negative will not work either since we will then get a greatest coefficient of 1458 rather than the correct answer of 1215. 58 Australian Senior Mathematics Journal 20 (1) Pahor [Discussion question: How could the above method be modified to properly deal with situations where d < 0?] We are now in a position to state and prove our central result. Theorem Suppose that a and b are positive real numbers and n is a positive integer. Then the largest coefficient in the binomial expansion of (ax + b)n is Proof All coefficients are multiplied by bn so by the above lemma, the largest coeffi-cient is given by where as required. QED. Returning to our original HSC question regarding the expansion of (3x + 7)25 we have a = 3, b = 7, and n = 25. Thus The largest coefficient is therefore [Investigation: Try to establish a formula for the greatest coefficient in the expansion of (axp + bxq)n. What restrictions do you need to place on a, b, p and q?] 59 Largest coefficients in binomial expansions Australian Senior Mathematics Journal 20 (1)
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https://pmc.ncbi.nlm.nih.gov/articles/PMC5018209/
CANCER OF THE ORAL CAVITY - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer | PMC Copyright Notice Surg Oncol Clin N Am . Author manuscript; available in PMC: 2016 Sep 10. Published in final edited form as: Surg Oncol Clin N Am. 2015 Apr 15;24(3):491–508. doi: 10.1016/j.soc.2015.03.006 Search in PMC Search in PubMed View in NLM Catalog Add to search CANCER OF THE ORAL CAVITY Pablo H Montero Pablo H Montero, MD Head and Neck Surgery Service, Department of Surgery, Memorial Sloan-Kettering Cancer Center Find articles by Pablo H Montero , Snehal G Patel Snehal G Patel, MD Head and Neck Surgery Service, Department of Surgery, Memorial Sloan-Kettering Cancer Center Find articles by Snehal G Patel Author information Article notes Copyright and License information Head and Neck Surgery Service, Department of Surgery, Memorial Sloan-Kettering Cancer Center ✉ Corresponding Author: Dr. Snehal G. Patel, Head and Neck Surgery, Memorial Sloan-Kettering Cancer Center, 1275 York Ave, New York, NY 10065. Telephone: 212-639-3412, patels1@mskcc.org Issue date 2015 Jul. Keywords: oral cavity cancer, oral cancer, squamous cell carcinoma, head and neck cancer PMC Copyright notice PMCID: PMC5018209 NIHMSID: NIHMS811187 PMID: 25979396 The publisher's version of this article is available at Surg Oncol Clin N Am INTRODUCTION Cancer of the oral cavity is one of the most common malignancies,1 especially in developing countries, but also in the developed world2. Squamous cell carcinoma (SCC) is the most common histology and the main etiological factors are tobacco and alcohol use3. Although early diagnosis is relatively easy, presentation with advanced disease is not uncommon. The standard of care is primary surgical resection with or without postoperative adjuvant therapy. Improvements in surgical techniques combined with the routine use of postoperative radiation or chemoradiation therapy have resulted in improved survival statistics over the past decade 4. Successful treatment of patients with oral cancer is predicated on multidisciplinary treatment strategies to maximize oncologic control and minimize impact of therapy on form and function. ANATOMY OF THE ORAL CAVITY The oral cavity extends from the vermilion border of the lips to the circumvallate papillae of the tongue inferiorly and the junction of the hard and soft palate superiorly. The oral cavity is divided into several anatomical subsites: lip, oral tongue, floor of mouth, buccal mucosa, upper and lower gum, retromolar trigone and hard palate (Figure 1). Despite their proximity, these subsites have distinct anatomical characteristics that need to be taken into account in planning oncologic therapy. Figure 1. Open in a new tab Anatomic sites of the oral cavity From Shah JP, Patel SG, Singh B, et al. Jatin Shah's head and neck surgery and oncology. 4th ed. Philadelphia, PA: Elsevier/Mosby; 2012, 232–244 with permisison. EPIDEMIOLOGY AND ETIOLOGY Worldwide, 405,000 new cases of oral cancer are anticipated each year, and the countries with the highest rates are Sri Lanka, India, Pakistan, Bangladesh, Hungary and France5 (Figure 2). In the European Union there are an estimated 66,650 new cases each year. The American Cancer Society estimates that there will be 42,440 new cancers of the oral cavity and pharynx in the U.S. causing 8,390 deaths in 20146. Tobacco smoking and alcohol are the main etiological factors in SCC of the oral cavity (SCCOC)3, 7. Other habits such as betel nut and tobacco chewing have been implicated in the Asian population. Figure 2. Open in a new tab Incidence of oral cavity cancer among both sexes expressed by level of Age-standardized rate (ASR) in countries of the world (From GLOBOCAN 2012 International Agency for Research on Cancer ( Tobacco contains many carcinogenic molecules, especially polycyclic hydrocarbons and nitrosamines. A directly proportional effect exists between the pack years of tobacco used and the risk of SCCOC8. This risk can be reduced after tobacco cessation, but it does not fully abate (30% in the first 9 years and 50% for those over 9 years)9,10. A decreased incidence of oral cavity cancer has been reported in the last 15 years, widely attributed to a reduction in tobacco use11. Alcohol and tobacco seem to have a synergistic effect in the etiology of oral and oropharyngeal SCC3, 12, 13. However, alcohol is linked to an increased risk of cancer even in non-smokers14. Other factors such as poor oral hygiene 15, wood dust exposure 16, dietary deficiencies17, red meat and salted meat consumption18, 19 have been reported as etiologic factors. The herpes simplex virus (HSV) has been suspected but has not been implicated in the etiology of SCCOC20. Despite the emerging evidence supporting the role of the human papilloma virus (HPV) in the etiology of oropharyngeal cancer, it has not been conclusively linked to SCCOC21. Host factors such as immune system alterations in transplant patients22, 23 and HIV-infected patients with AIDS24, and genetic conditions like xeroderma pigmentosum, Fanconi anemia and ataxia telangiectasia are associated with an increased incidence of head and neck cancer25–28. Oral cancer is more common in men and usually occurs after the 5 th decade of life. About 1.5% will have another synchronous primary in the oral cavity or the aero-digestive tract (larynx, esophagus or lung)29. Metachronous tumors develop in 10% to 40% in the first decade after treatment of the index primary30, 31 and therefore regular post-therapy surveillance and lifestyle alteration are important strategies for secondary prevention. PATHOLOGY Squamous cell carcinomas (SCC) constitute more than 90% of all oral cancer. Other malignant tumors can arise from the epithelium, connective tissue, minor salivary glands, lymphoid tissue, and melanocytes or metastasis from a distant tumor. A variety of premalignant lesions have been associated with development of SCC32. The more common premalignant lesions including leukoplakia, erythroplakia, oral lichen planus, and oral submucous fibrosis have varying potential for malignant transformation33. The WHO (2005) classifies premalignant lesions according to degree of dysplasia into mild, moderate, severe, and carcinoma in situ. Leukoplakia is a clinical term defined as a “white patch or plaque that cannot be characterized clinically or pathologically as any other disease”34. This lesion is usually associated with smoking and alcohol use. The prevalence of leukoplakia worldwide is about 2%. Dysplastic changes are seen in only 2–5% of patients. The annual rate of malignant transformation for leukoplakia is 1%. Risk factors for malignant transformation include presence of dysplasia, female gender, long duration of leukoplakia, location on the tongue or floor of mouth, leukoplakia in non-smokers, size greater than 2cm, and non-homogeneous type. In addition to lifestyle alteration to avoid tobacco and alcohol use, excision constitutes the only definitive modality for accurate diagnosis and treatment. Erythroplakia is a “bright red velvety patch that cannot be characterized clinically or pathologically as being caused by any other condition”34. Surgical excision is recommended as these lesions have higher malignant potential than leukoplakia and are commonly associated with dysplasia and carcinoma in situ. Non-squamous cell carcinomas of the oral cavity are uncommon. Minor salivary gland carcinomas represent less than 5% of the oral cavity cancers. They frequently arise on the hard palate (60%), lips (25%) and buccal mucosa (15%)35. Mucoepidermoid carcinoma is the most common type (54%), followed by low-grade adenocarcinoma (17%), and adenoid cystic carcinoma (15%)49,50. Mucosal melanomas are rare but usually present as locally aggressive tumors, mainly of the hard palate and gingiva. Bony tumors including osteosarcoma of the mandible or maxilla and odontogenic tumors such as ameloblastoma can present within the oral cavity and may be mistaken for a mucosal lesion if there is surface ulceration. CLINICAL PRESENTATION AND EVALUATION Despite easy self-examination and physical examination, patients often present with advanced stage disease. A comprehensive head and neck exam is mandatory in patients with suspected oral cavity cancer. Visual inspection and palpation allow an accurate impression of the extent of the disease, the third dimension of tumor, the presence of bone invasion, or skin breakdown. Appropriate documentation with drawings and photographic records of the tumor are useful in staging, decision-making and further follow up. The clinical TNM stage should be recorded at first encounter and modified as evaluation progresses. The initial workup consists of diagnosis by biopsy. Accessible lesions may be adequately biopsied in the clinic using punch forceps, core needle or fine-needle aspiration. Some patients will require examination under general anesthesia (EUA) in order to access posteriorly located lesions, or to complete a physical exam limited by pain and trismus. Radiographic imaging is crucial for evaluation of the relation of the tumor to adjacent bone and for assessing regional lymph nodes. CT scan is the study of choice for evaluation of bone and neck nodes, especially early cortical involvement and extracapsular nodal spread. MRI provides complementary information about soft tissue extent and perineural invasion and is also helpful for evaluating the extent of medullary bone involvement because adult marrow is normally replaced by fat. Most patients with oral cancer are not at risk for distant metastases and therefore the role of PET scan in initial assessment is debatable. However, a preoperative PET scan may be useful as a baseline if adjuvant treatment is anticipated and a PET scan will be used for radiation therapy planning (though this is undertaken differently from a “diagnostic” PET scan). Patients with locally advanced tumors require appropriate multidisciplinary consultations with the reconstructive surgeon, medical specialists for presurgical optimization, dental professionals, speech and swallowing pathologists, and behavioral therapists for smoking cessation and other lifestyle alterations. The TNM system is the most widely accepted prognostic system due to its relatively simple design and user-friendliness. The clinical staging of the oral cavity tumors consists of primary tumor characteristics, the neck, and assessment for distant metastases (Table 1). This information allows TNM stage grouping for the tumor (Table 2)36. The basic elements in staging of the primary site are the tumor size and invasion of deep structures. Advanced disease is defined by invasion of structures such as medullary bone, deep muscle of the tongue, maxillary sinus, and skin for T4a disease, or masticator space, pterygoid plates, or skull base and/or encasement of the internal carotid artery for T4b disease. Lymphatic spread into the neck generally occurs in a step-wise, orderly and predictable fashion. The lymph node echelons of the neck are described using the terminology standardized by the American Head and Neck Society Guidelines37 (Figure 3). Table 1. TNM classification of carcinomas of the oral cavity | T — Primary tumor | | TX | Primary tumor cannot be assessed | | T0 | No evidence of primary tumor | | Tis | Carcinoma in situ | | T1 | Tumor 2 cm or less in greatest dimension | | T2 | Tumor more than 2 cm but not more than 4 cm in greatest dimension | | T3 | Tumor more than 4 cm in greatest dimension | | T4a (lip) | Tumor invades through cortical bone, inferior alveolar nerve, floor of mouth, or skin (chin or nose) | | T4a (oral cavity) | Tumor invades through cortical bone, into deep/extrinsic muscle of tongue (genioglossus, hyoglossus, palatoglossus, and styloglossus), maxillary sinus, or skin of face | | T4b (lip and oral cavity) | Tumor invades masticator space, pterygoid plates, or skull base; or encases internal carotid artery | | Note: Superficial erosion alone of bone/tooth socket by gingival primary is not sufficient to classify a as T4. | | N - Regional Lymph Nodes | | NX | Regional lymph nodes cannot be assessed | | N0 | No regional lymph node metastasis | | N1 | Metastasis in a single ipsilateral lymph node, 3 cm or less in greatest dimension | | N2 | Metastasis as specified in N2a, 2b, 2c below | | N2a | Metastasis in a single ipsilateral lymph node, more than 3 cm but not more than 6 cm in greatest dimension | | N2b | Metastasis in multiple ipsilateral lymph nodes, none more than 6 cm in greatest dimension | | N2c | Metastasis in bilateral or contralateral lymph nodes, none more than 6 cm in greatest dimension | | N3 | Metastasis in a lymph node more than 6 cm in greatest dimension | | Note: Midline nodes are considered ipsilateral nodes. | | M – Distant metastasis | | MX | Distant metastasis cannot be assessed | | M0 | No distant metastasis | | M1 | Distant metastasis | Open in a new tab From, Edge SB, Byrd DR, Compton CC, eds. AJCC Cancer Staging Manual. 7th ed. New York, NY.: Springer, 2010; 33, with permission. Table 2. Oral cancer staging | Stage | T | N | M | :--- :--- | | 0 | Tis | N0 | M0 | | I | T1 | N0 | M0 | | II | T2 | N0 | M0 | | III | T3 | N0 | M0 | | T1 | N1 | M0 | | T2 | N1 | M0 | | T3 | N1 | M0 | | IVA | T4a | N0 | M0 | | T4a | N1 | M0 | | T1 | N2 | M0 | | T2 | N2 | M0 | | T3 | N2 | M0 | | T4a | N2 | M0 | | IVB | Any T | N3 | M0 | | T4b | Any N | M0 | | IVC | Any T | Any N | M1 | Open in a new tab From, Edge SB, Byrd DR, Compton CC, eds. AJCC Cancer Staging Manual. 7th ed. New York, NY.: Springer, 2010; 33, with permission. Figure 3. Open in a new tab Cervical lymph node level classification From Shah JP, Patel SG, Singh B, et al. Jatin Shah's head and neck surgery and oncology. 4th ed. Philadelphia, PA: Elsevier/Mosby; 2012, 232–244, with permisison. Knowledge of the patterns of nodal metastasis has practical implications in the design of neck dissection for patients with oral cancer. The patient with a clinically negative neck is at highest risk of metastasis to levels I–III38. Skip metastases to level IV do occur, especially in cancer of the anterior tongue. Metastases to level V are extremely rare (1%) even in patients with clinically positive neck. Oral tongue tumors have the greatest propensity of all oral cancers for metastasis to the neck, and tumor thickness (Figure 4) is a major predictor of risk of nodal metastasis39. Figure 4. Open in a new tab Incidence of lymph node metastasis and survival stratified by the thickness of the primary tumor. (From Shah JP, Patel SG, Singh B, et al. Jatin Shah's head and neck surgery and oncology. 4th ed. Philadelphia, PA: Elsevier/Mosby; 2012, 232–244, with permisison.) TREATMENT Surgical resection is the treatment of choice for SCCOC. Surgical resection allows accurate pathologic staging, with information about the status of margins, tumor spread and histopathologic characteristics which can then be used to inform subsequent management based upon assessment of risk versus benefit. Adjuvant radiotherapy ± chemotherapy is used for specific indications in locoregionally advanced tumors. A multidisciplinary team is absolutely essential to ensure a favorable outcome. Multiple factors are taken into account in selecting treatment for an individual patient. The risk of treatment-related complications should be assessed based on physiological age, comorbid conditions (e.g. cardiopulmonary status), lifestyle (smoking or alcohol), surgical resectability, and patient expectations. Surgical Management A detailed description of surgical technique for management of oral cavity cancers is beyond the scope of this publication and the reader is referred to specialized texts for this information40. Broad principles of surgical management will be discussed and these include access to the oral cavity, management of the mandible, management of neck nodes, and reconstruction of oral cavity surgical defects. Surgical access The transoral approach is usually used for premalignant lesions and small, superficial tumors of the anterior floor of mouth, alveolus and tongue. A more invasive approach becomes necessary for posteriorly located tumors or if there are limitations due to trismus or inadequate surgical exposure (Figure 5). The lip-splitting paramedian mandibulotomy approach is used for larger posteriorly located tumors of the tongue. The upper cheek flap and midfacial degloving approaches are useful for gaining access to the maxilla. Figure 5. Open in a new tab Various surgical approaches. A, Peroral. B, Mandibulotomy. C, Lower cheek flap. D, Visor flap. E, Upper cheek flap. (From Shah JP, Patel SG, Singh B, et al. Jatin Shah's head and neck surgery and oncology. 4th ed. Philadelphia, PA: Elsevier/Mosby; 2012, 232–244, with permisison.) Management of the mandible Mandibular invasion can occur early in tumors of the floor of the mouth, the ventral surface of the tongue and the gingivobuccal sulcus. The mechanism of invasion of these tumors into the mandible has been well studied41–43. Tumors invade the mandible through the dental sockets in the dentate mandible, and through the dental pores of the alveolar process in the edentulous mandible. Early cortical invasion of the mandible is difficult to assess with plain radiography, or orthopantomograms but CT scans are more sensitive. On a practical basis, tumors that are in close juxtaposition to the mandibular cortex will require consideration for marginal mandibulectomy in order to achieve an adequate margin of resection irrespective of radiographically demonstrable early cortical invasion. The role of marginal resection might be limited in patients with reduced vertical height of the body of the mandible due to the higher risk of early involvement of the body of mandible and the risk of pathologic fracture if a marginal resection is performed. Adequate tumor clearance in edentulous patients may therefore necessitate a segmental mandibulectomy. The indications for segmental resection are listed in Table 3. Table 3. Indications for Segmental mandibulectomy Gross invasion of the of the mandible Tumor fixation to the majority of the vertical height of the occlusal surface of the mandible in hypoplastic edentulous mandible with significant loss of vertical height precluding safe performance of rim resection Tumor fixed to the mandible following prior radiotherapy to the mandible Open in a new tab Management of the neck Sixty percent of patients with early stage oral cancer will present with a clinically negative neck (cN0). Approximately 20–30% will have microscopically evident nodal metastasis on histologic examination after elective neck dissection (END). The risk of nodal metastasis is related to several factors (Table 4)44, 45. Cervical lymph node metastasis is the single most important prognostic factor in oral cancer: survival chances are reduced by 50% when compared to those with similar primary tumors without neck metastases46, 47. SCC of the oral tongue and the floor of the mouth are more likely to metastasize to the neck, and these patients should be offered END, even for early stage tumors, if they are thicker than about 4mm48. The hard palate and the upper gum have a relatively lower rate of occult nodal metastasis and END may not be indicated49. Table 4. Risk factors of nodal metastasis in oral cancer Tumor Size Histologic Grade Depth of Invasion Perineural Invasion Vascular Invasion Open in a new tab Sentinel node biopsy is an alternative to END for staging the cN0 neck in early stage (T1–2) SCCOC. The technique was first reported in 2001 by Shoaib et al50 and has been analyzed in several single institutional studies as well as two prospective multicenter trials, one in Europe51, 52 and the other in the US53. The procedure is technically challenging and successful identification of sentinel nodes and detecting occult metastasis depends on expertise and experience. Therefore, it should be undertaken only in centers with the necessary proficiency and the appropriate volume of cases54. In patients with clinically or radiographically involved neck nodes, a therapeutic comprehensive neck dissection is indicated (Table 5). It involves dissection of levels I to V. The need to sacrifice other structures such as the spinal accessory nerve, sternocleidomastoid muscle, or internal jugular vein depends on the location of the metastasis and its characteristics. The most common type of comprehensive neck dissection is the modified radical neck dissection, MRND Type 1. Radical neck dissection is rarely performed unless there is direct infiltration of the relevant structures by gross extranodal extension of disease (Table 5). Table 5. Types of Neck Dissections | | Lymph Nodes Excised | Other Structures Excised | Structures Preserved | :--- :--- | | Radical Neck Dissection (RND) | Levels I–V | Sternocleidomast oid Muscle, Internal Jugular Vein, Spinal Accessory Nerve, Submandibular Gland | | | Modified Radical Neck Dissection (MRND) Type I | Levels I–V | Sternocleidomast oid Muscle, Internal Jugular Vein, Submandibular Gland | Spinal Accessory Nerve | | Modified Radical Neck Dissection (MRND) Type II | Levels I–V | Internal Jugular Vein, Submandibular Gland | Sternocleidomoid Muscle, Spinal Accessory Nerve | | Modified Radical Neck Dissection (MRND) Type III | Levels I–V | Submandibular Gland | Sternocleidomast oid Muscle, Internal Jugular Vein, Spinal Accessory Nerve | | Supraomohyoid Neck Dissection (SOHND) | Levels I–III | Submandibular Gland | Sternocleidomast oid Muscle, Internal Jugular Vein, Spinal Accessory Nerve | Open in a new tab In a patient with a clinically negative neck, the risk of occult metastasis is mainly to levels I through III. Potential compromise of levels IV and V is very rare. For these reasons, a supraomohyoid neck dissection (SOHND)(Table 5) is usually adequate to stage the cN0 neck. In patients with primary oral tongue SCCOC dissection of level IV may be indicated due to the possibility of skip metastasis. For patient with positive nodes on END, neck recurrence is observed in 10–24%55. Appropriately selected patients benefit from postoperative radiation therapy56, 57. For cN 0 patients who are proven pathologically N 0, failure rates of less than 10% have been reported58. Reconstructive surgery Restoration of form and function after ablative cancer surgery is the ultimate goal of treatment and is achieved by choosing the appropriate reconstructive procedure. Surgical defects after resection of early stage tumors can usually be reconstructed with primary closure or the use of skin graft or skin substitutes. Reconstruction of larger and more complex defects that result from resection of advanced tumors requires participation from an expert reconstructive surgeon. Microvascular free tissue transfer is the technique of choice59, 60. For example, in patients with soft tissue defects of the oral tongue, floor of mouth and retromolar trigone, the free radial forearm flap results in excellent functional results (Figure 6). In addition to soft tissue cover, free flaps are also a reliable source for bone reconstruction. The fibula free flap is currently the workhorse in reconstruction of defects following segmental mandibulectomy (Figure 6). Other composite microvascular flaps include the radial forearm osteocutaneous flap, iliac crest and scapula free flaps. Several studies have demonstrated the reliability and low morbidity of microvascular free flap reconstruction techniques61. The ability reliably to reconstruct large surgical defects has contributed to improved oncologic outcomes in patients with locally advanced cancers by enabling more complete resections62. Pedicled myocutaneous flaps such as the pectoralis major, latissimus dorsi or trapezius flaps are reliable alternatives if surgical expertise is not available or if the patient is not a good candidate for microvascular reconstruction. Figure 6. Open in a new tab Fibular (left) and radial forearm (right) free flaps are two of the most common flaps used in oral cavity reconstruction after major resections. Adjuvant treatment Adjuvant postoperative treatment is indicated in patients with high risk of locoregional recurrence. This includes patients with large primary tumors (pT3 or pT4), bulky nodal disease (pN2 or pN3), metastases to nodal levels IV or V, positive surgical margins, lymphovascular invasion, perineural invasion, and extracapsular spread. External beam radiation therapy has been the traditional modality for postoperative adjuvant treatment and doses of 66–70 Gy result in good locoregional control63, 64. Two clinical trials have shown that administration of cisplatin chemotherapy concurrently with postoperative radiotherapy improves locoregional control and survival (versus radiotherapy alone) in head and neck cancer patients with extracapsular spread and /or positive surgical margins65, 66. However, concurrent chemoradiation can result in significant morbidity and is best used at centers where appropriate expertise and infrastructure is available. OUTCOMES OF TREATMENT The results of treatment of SCCOC in recently published major series are shown in Table 6. The overall 5-year survival in a recently analyzed cohort of patients at Memorial Sloan-Kettering Cancer Center is 63%. This represents a significant improvement compared to historical cohorts (Figure 7) and may be related to wider use of microvascular free flaps with enhanced ability to resect large tumors and reconstruct large and complex defects, more aggressive regional therapy including increasing use of elective selective neck dissections, and the use of postoperative adjuvant therapy. Table 6. Outcomes in patients treated for squamous carcinoma of the oral cavity in major series around the world. | Series | Country | Year | Total no. of Patients | 5-years OS all patients | 5-years DSS all patients | Stage I | Stage II | Stage III | Stage IV | :--- :--- :---: :---: :---: | | Loree et al72 | USA | 1997 | 398 | 57.0% - - | Chen et al73 | Taiwan | 1999 | 7032 | 36.1% 72.0% | 38.9% | 26.7% | 11.8% | | Funk et al74 | USA | 2000 | 30,803 | 43.5% - - | Carvalho et al75 | Brazil | 2002 | 3642 | 43.0% 74.0% | 33.0% | | Yeole et al76 | India | 2000 | 15051 | 45.9% 68.9% | 26.6% | 9.5% | | Rogers et al77 | UK | 2008 | 541 | 56.0% | 74.0% - - | | Listl et al78 | Germany | 2012 | 15792 | 54.6% - - | MSKCC | USA | 2014 | 1816 | 62.5% 78.5% | 68.4% | 64.5% | 34.5% | Open in a new tab Figure 7. Open in a new tab Outcomes of treatment of SCCOC in three cohorts treated during different time periods at Memorial Sloan-Kettering Cancer Center (1960–2005). Courtesy of Memorial Sloan-Kettering database, New York, NY. Approximately a third of patients treated for SCCOC relapse, and locoregional recurrence is the most common pattern of failure. The clinical stage at presentation is an important predictor of survival (Figure 8) but the most powerful predictor of outcome is the presence of metastatic lymph nodes (Figure 9). Other clinical signs of locally advanced disease and poor prognosis include trismus, which indicates invasion of the pterygoid, temporalis or masseter muscle; reduced tongue mobility, which indicates invasion of the extrinsic musculature of the tongue or the hypoglossal nerve; and skin invasion with dermal lymphatic infiltration. Significant histopathologic predictors of outcome include depth of invasion of the primary tumor, positive margins of surgical resection, perineural invasion and major extracapsular nodal extension. Figure 8. Open in a new tab Clinical stage at presentation is an important predictor of outcome. Courtesy of Memorial Sloan-Kettering database, New York, NY. Figure 9. Open in a new tab Impact of clinically palpable lymph node metastasis on disease-specific survival in SCCOC. Courtesy of Memorial Sloan-Kettering database, New York, NY. Follow up Oral cancer patients have a high risk of locoregional recurrence and developing subsequent new primary cancers, but the risk of distant recurrence is low67. The possibility of a second head and neck primary is about 4–7% a year and comprehensive clinical examination and a high suspicion are the cornerstones of early diagnosis68. Control of lifestyle-related risk factors, such as tobacco and alcohol consumption, is a priority in these patients because of the higher risk of treatment failure and second primaries69. Unfortunately, there is no effective chemoprevention and close follow up remains the most important tool in secondary prevention70. Baseline imaging studies are often obtained about 3–6 months following completion of treatment and then as needed based on clinical suspicion. Chest imaging is not routinely needed but may be beneficial in patients with a significant smoking history. Other ancillary measures include speech and swallowing rehabilitation as indicated, monitoring of thyroid stimulating hormone levels if the neck been treated with radiation therapy, and regular dental evaluation. CONCLUSION Treatment results for patients with oral cancer have improved considerably over the last several decades due to improvements in reconstruction and adjuvant treatment. Further improvements in survival have been hampered by attrition from second and subsequent primary tumors in long-term survivors. Primary and secondary prevention of oral cancer requires better education about lifestyle related risk factors, and improved awareness and tools for early diagnosis. Key Points. Cancer of the oral cavity is a common malignancy in the United States and around the world. The standard of care is primary surgical resection with or without postoperative adjuvant therapy. Multidisciplinary treatment is crucial to improve the oncologic and functional results in oral cancer patients Primary and secondary prevention of oral cancer requires education about lifestyle-related risk factors, and improved awareness and tools for early diagnosis. Footnotes The Authors have nothing to disclose. Publisher's Disclaimer: This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final citable form. 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[DOI] [PMC free article] [PubMed] [Google Scholar] ACTIONS View on publisher site PDF (1.1 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page INTRODUCTION ANATOMY OF THE ORAL CAVITY EPIDEMIOLOGY AND ETIOLOGY PATHOLOGY CLINICAL PRESENTATION AND EVALUATION TREATMENT OUTCOMES OF TREATMENT CONCLUSION Footnotes References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
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Daily Crossword Word Puzzle Word Finder All games Word of the Day Word of the Year New words Language stories All featured Slang Emoji Memes Acronyms Gender and sexuality All culture Writing tips Writing hub Grammar essentials Commonly confused All writing tips Games Featured Culture Writing tips Advertisement View synonyms for electorate electorate [ih-lek-ter-it] noun the body of persons entitled to vote in an election. the dignity or territory of an Elector of the Holy Roman Empire. electorate / ɪˈlɛktərɪt/ noun the body of all qualified voters the rank, position, or territory of an elector of the Holy Roman Empire the area represented by a Member of Parliament the voters in a constituency Discover More Word History and Origins Origin ofelectorate1 First recorded in 1665–75; elector + -ate 3 Discover More Example Sentences Examples are provided to illustrate real-world usage of words in context. Any opinions expressed do not reflect the views of Dictionary.com. The electorate has shown politicians they can move in droves, and fast. FromBBC It’s an off-year election, which means turnout is likely to be low and the electorate is unpredictable. FromLos Angeles Times "We have an avoidable habit of not listening enough to the electorate," he says. FromBBC The most conscientious pollsters invest considerable time and effort figuring out how to model their voter samples — that is, how to best reflect the eventual composition of the electorate. FromLos Angeles Times But some Labour figures wonder if as many as 80 more could choose not to officially back anyone, shrivelling the electorate. FromBBC Advertisement Discover More Related Words constituency voter Advertisement Advertisement Advertisement electoral voteelectr-
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IB Colourful Solutions in Chemistry Colourful Solutions - IB Chemistry Table Data Book Calculator Next page Colourful Solutions>Ideal gases> The gas laws Your browser does not support HTML5 video. IB Chemistry Shop Standard level If the number of moles present in a gas and its mass is known, then the relative molecular mass of a gas can be calculated. Syllabus ref: S1.5.4 Structure 1.5.4 - The relationship between the pressure, volume, temperature and amount of an ideal gas is shown in the ideal gas equation PV = nRT and the combined gas law P 1 V 1/T 1= P 2 V 2/T 2. Solve problems relating to the ideal gas equation. Guidance Units of volume and pressure should be SI only. The value of the gas constant R, the ideal gas equation, and the combined gas law, are given in the data booklet. Tools and links Tool 1, Inquiry 2 - How can the ideal gas law be used to calculate the molar mass of a gas from experimental data? Relative molecular mass calculation Worked examples Now test yourself Ideal gas equation The equation of state refers to a fixed mass of gas. From Avogadro's law we know that the same volume of all gases contain the same number of moles and from this, it follows that the volume is proportional to the number of moles. Volume ∝ number of moles (n) These two equations can be combined to obtain an expression involving all the quantities: After rearrangement, for 'n' moles of gas the proportionality constant is called the Universal Gas Constant and is given the symbol 'R' This gives the ideal gas equation: Ideal Gas Equation: PV = nRT where: P = pressure in Pa V = volume in m 3 n = number of moles of gas R = Universal Gas constant = 8.314 JK-1 mol-1 T = the absolute temperature in Kelvin It is often more convenient to express the pressure in kPa and the volume in litres (dm 3). This leaves the value of R the same (see below). Example: Calculate the number of moles of gas present in 2.6 dm 3 at a pressure of 1.01 x 10 5 Pa and 300 K. PV = nRT 2.6 dm 3 = 0.0026 m 3 0.0026 x 1.01 x 10 5 = n x 8.314 x 300 n = 0.0026 x 1.01 x 10 5 / 8.314 x 300 n = 0.105 moles There are several units used for gas volume, gas pressure and temperature. It is important to be consistent with the use of units when carrying out gas law calculations. The Syllabus states that SI units will be used wherever possible. ^ top Universal gas constant - R Although called "Universal", its value depends on the units used for P, V and T. With the SI units of metres, kilograms, Kelvin and Joules, using P, V and T values at STP gives: PV=nRT therefore:R=PV/nT for 1 mole of gas at STP (using accepted values of P = 1.00 x 10 5 Pa, V = 0.02271 m 3, T = 273.15 K) R =(1.00 x 10 5) x 0.02271)/273.15 R =8.314 J K-1 mol-1 In chemistry, the units of volume used are the decimetre cubed (dm 3) and pressure in kiloPascals (kPa), so one unit is 100x greater and the other 100x smaller than the SI equivalent. Consequently the differences in the product, PV, both cancel out (multiplying AND dividing by 1000), so that the final value for R is the same as in SI units. The Universal gas constant, R, calculated using atmospheres Pressure and volume in litres, then: PV=nRT R=PV/nT at STP:P = 1 atm, V = 22.7 dm 3, T = 273 n = 1 R = 0.0821 dm 3 atm mol-1 K-1 There are, of course, several other values of R, as there are several ways of measuring both the volume and the pressure of a gas. ^ top SI units and 'R' The SI units of P, V and T give rise to the previously used value for the universal gas constant, R = 8.314 J K-1 mol-1. How does this happen when chemists do not use these SI units? Remember: 1 litre = 1 dm 3 = 1000 cm 3 Consequently, if litres are used in the Ideal Gas equation then the pressure units must also be divided by 1000 (as PV = constant). Pressure is measured in Pa or Nm-1, and so the unit of the kPa correct for the difference in volume units. Atmospheric pressure in Pa = 1.00 x 10 5 Pa Atmospheric pressure in kPa = 1.00 x 10 2 kPa Provided that you are consistent with the application of units there will be no problem. It is always a good idea when carrying out calculations to look at the value of your answer and ask yourself, "does it seem reasonable?" ^ top Worked examples Q154-01 What volume is needed to store 50 moles of an ideal gas at 15 atmospheres and 25 ºC? Answer PV = nRT Convert pressure to kPa and temperature to Kelvin. P = (15 x 100) = 1500 kPa, n = 50 mol, T = (273 + 25) = 298 K, R = 8.314 J K-1 mol-1 V = nRT/P = (50 x 8.314 x 298)/1500 Therefore volume needed = 82.6 dm 3 Q154-02 What pressure will be exerted by 200 moles of hydrogen gas in a 7.5L cylinder at 20ºC? Answer PV = nRT Convert temperature to Kelvin. V = 7.5 dm 3, n = 200 moles, T = (273 + 20) = 293 K, P = nRT/V Pressure exerted = (200 x 8.314 x 293)/7.5 Therefore pressure exerted = 64960 kPa Q154-03 The temperature in Kelvin of 2 dm 3 of an ideal gas is doubled and its pressure is increased by a factor of 4. Calculate the final volume of the gas. Answer The ideal gas equation: PV = nRT Shows that when the temperature is doubled so the value of the volume must double also, ie a twofold increase (initial volume x 2). However, when nRT is constant, a fourfold increase in pressure results in a fourfold (x4) decrease in volume (= initial volume/4) Combining the two effects (that of the temperature and the pressure change) the volume increases by a factor of 2 and decreases by a factor of 4 (volume = initial volume x 2/4) Therefore the overall change in gas volume is a twofold decrease (volume = initial volume/2) Final volume is therefore 2/2 = 1 dm 3 Q154-04 Calculate the volume occupied by 0.01 moles of hydrogen gas at 20ºC and atmospheric pressure. Answer Gas molar volume at STP = 22.7 dm 3 0.01 moles of gas occupies 0.01 x 22.7 = 0.227 dm 3 20ºC is equivalent to 20 + 273 K = 293 K Using V 1/T 1 = V 2/T 2 (where V 1 and T 1 are under STP conditions) 0.227/273 = V 2/293 Therefore V 2 = (0.227 x 293)/273 = 0.2436 dm 3 = 243.6 cm 3 Therefore 0.01 moles of hydrogen occupies 243.6 cm 3 at 20ºC Q154-05 How many moles of gas are present in a gas volume of 24 dm 3 at 100ºC and atmospheric pressure? Answer Using the ideal gas equation PV=nRT 100ºC = 373K, R = 8.314, atmospheric pressure = 100 kPa n = PV/RT= (100 x 24)/(8.314 x 373) n = 0.774 moles Q154-06 For which set of conditions does a fixed mass of an ideal gas have the greatest volume? Temperature Pressure A low low B low high C high high D high low Answer From the ideal gas equation, PV = nRT When T is high, V must also be high but as nRT is equal to a constant value, any increase in P must cause a decrease in V Therefore V is high when P is low. Correct response = D Q154-07 A 0.450g sample of gaseous aluminium chloride occupies a volume of 51.2 cm 3 at 100ºC and 102 kPa. Calculate its relative molecular mass. Answer 100ºC is equivalent to 373K, 51.2 cm 3 = 0.0512 dm 3 Using PV=nRT, n = PV/RT n = (102 x 0.0512)/(8.314 x 373) n = 1.684 x 10-3 The mass of 1.684 x 10-3 moles = 0.450 g Therefore the mass of 1 mole = 0.450/1.684 x 10-3 = 267 The relative molecular mass of aluminium chloride = 267 Note: The formula of aluminium chloride is AlCl 3, however this gives a relative formula mass of 27 +(3 x 35.5) = 133.5 exactly half the calculated value. The Mr value produced in the above calculation suggests that the correct formula of aluminium chloride is 2 x AlCl 3 or Al 2 Cl 6 Q154-08 Calculate the number of moles of helium in a weather balloon measuring 200 m3 at 5ºC and 80 kPa pressure. Answer Using PV=nRT n=PV/RT Volume = 200 m 3 = 2 x 10 5 dm 3 Number of moles of helium = (80 x 2 x 10 5 )/(8.314 x 278) Therefore the number of moles of helium = 6923 moles (to 4 significant figures) Q154-09 The molar mass of an unknown gas is to be determined by weighing a sample. As well as its mass which of the following must be known. 1. I. Pressure 2. II. Temperature 3. III. Volume I only II only I and II only I, II, and III Answer To calculate the molar mass of a gas the number of moles must be known as well as the mass of gas to use the relationship: Molar mass = Mass/moles The number of moles is obtained from PV=nRT Therefore the Pressure, Volume and Temperature must be known. Correct response = D Q154-10 In an experiment to determine the relative molecular mass of an unknown hydrocarbon, 0.15 g of the liquid hydrocarbon was injected through a rubber septum into a gas syringe. The gas syringe was placed in an oven at 120ºC and left to reach equilibrium. The final volume occupied by the vapour in the syringe was measured and found to be 67.4 cm 3. Calculate the relative molecular mass of the hydrocarbon, if the atmospheric pressure was 101 kPa on that day. Answer Temperature 120ºC is equal to 120 + 273 = 393K Volume 67.4 cm 3 = 0.0674 dm 3 Using PV = nRT n = PV/RT n = (101 x 0.0674)/(8.314 x 393) = 2.083 x 10-3 This number of moles has a mass = 0.15 g Therefore mass of 1 mole = 0.15/ (2.083 x 10-3) = 72 Therefore the relative molecular mass of the hydrocarbon = 72 ^ top Now test yourself Exercise 154 - The ideal gas equation. Take care with the units given. Give your answer to three significant figures. Your answer Questions Correct ^ top External video resources Combined gas law problems: Richard Thornley previous Colourful Solutions 2025 next page ×
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https://arxiv.org/pdf/2310.02637
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Iű;ÈÅ3+tÑ8Þ6 !šÈµBÜ\iõµ2Ðð=åã¥o”í½YÁ8Á9mÃÐNÄ3 =b‰¡bú#5Éu¿ÙñÁëCÜ·†xÖjS°±<-eÎ£„ãî“έh“íÃL·ùÔº4_ðÙà_ßê,ÿºwØ:‡ õɨ©®%!}Ä–Æ øLOі¦º5ù3DžYåƧL¯Çúnš$îý; H¾mñ¡è´¸ÁŽ†Ôwñ;ÅSƒ/µ2•4+ތv´+¿˜gü9¡¦\aLµX> stream xڍTT”]·¦K鐔¥sH‘îNi€af†n‘îI$$i¤¥[:Dé¾£~ÿÿÝï¿w­{׬õÎ9ϳ÷>{ŸýìÃÆ¢kÀ'g·+Ãa(> ¿ $@AKÿ@PP˜\_PPˆˆÍ‚‚‚ÿ DlƇIþ7^¡Ð˜"…6ӂÃ�êîP�P�“ŠK „%þeGHA;€?@#‰Øà®Þˆƒ# }Ê¿–�N[.�PBBœ÷·;@Ό€Ø‚�-Êì‚>ÑÀm!”÷?Bp>tD¡\%<==ùA.H~8á/À‚r胑„Øð«\€6Èü»0~"6€¡#ù6€Û£•Zý$ºžß (©eN§þš­ºŠ€J©Jpz%ê èö]¤Üh¦3üìŒ58Uã@Ú,24åO¦1àùé~ÀvM¥Ñ¶Õ¡®[.7XttôN¾Ò¸ûK+—¡cbᰕ9©ÖsZNý/Þ©¼mH.Õ¥h©/.¹ïÈ|}¤šÕ>À¦Û$õ€Ãæ–@2Ö¬#µ-Ù¦Çn™Ø ²s¿¼~¿ñ0B"§ së:WB”¹ ’þwo„×Lˆq‰™´çÔldOKNɟý®"tÒÆmöBè[?_lo–k¯&Y5Ôw†=˜ozÈ¥UÞöå!–ËNl±ÔÀ×牌I¤åþ:v@K3ƒ!ª¯w]tCìE'¹˜^økEfêöm:TA±«ÊGC[$ßó\‰ÓŒ1êüÆøÖ=4'§f…¨ë˜–<œÅ¦0‰¢0ke º¿s¼Mú.Ž¹Ð@‹¿i–ƒÅøf6ux·ô5ŸQ¨bû¨‹ –ðPÍΛg;gt­ù «Å”ÆdÃ<ú3 W®9ފwU‡=¼|]ƒ(l5ipòN€žÍ›[Æ(è;©# 땷´´8 0£Vá2\8;2RͲZ”ù»òÚKe­&GñŠÒm816ì6±îۃæ&sŠ¸æÎߎczÌ÷PYí”èd÷?õÚ EÕçôÿ?9Ž¼äµk„±ÙÑÒ×rôÄ©À¶Ï¹æ¾»d=ž~ ÕLhDé«Q#YÓä«!²—wšWk'ù­LÙß½˜ cL€/2H ½&>ëïhjMl\'ªxàÔêC§6>·¥Æo¿,ËЁ=ø©ïp﹔bóó~†œ2¡DNužþ¿™Ñv•¬ä¢ÍÉݬoTËγUS!TïaÏ‰Ç §XþÙÓÛs�òÝo>“+ù Œõ׃ão荼¯ ]ÏÛPÓõq‘?¬/S…DŠ¿‚ÄƐÅIá_iXð&t2÷ Ó1÷«-–å\÷öz|NÛRÚò>ÖYo+æÆÿL‹”còêUWüì+°§›÷ éÕ÷^Í ? 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https://mathoverflow.net/questions/370302/a-claim-on-partitioning-a-convex-planar-region-into-congruent-pieces
mg.metric geometry - A claim on partitioning a convex planar region into congruent pieces - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community MathOverflow helpchat MathOverflow Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more A claim on partitioning a convex planar region into congruent pieces Ask Question Asked 5 years, 1 month ago Modified1 year, 3 months ago Viewed 456 times This question shows research effort; it is useful and clear 12 Save this question. Show activity on this post. Let us define a perfect congruent partition of a planar region R R as a partition of it with no portion left over into some finite number n of pieces that are all mutually congruent (ie any piece can be transformed into another piece by an isometry. We consider only cases where each piece is connected and is bounded by a simple curve). Note: It is known there are convex planar regions - indeed quadrilaterals - which do not allow perfect congruent partition for any n ( proves a stronger result). Claim: If a convex polygonal R R allows a perfect congruent partition of itself into N N non-convex pieces each with finitely many sides, then R R also allows a perfect congruent partition into N N convex pieces with finitely many sides. In other words, allowing the pieces to be non-convex polygons does not improve the chances of a convex planar region achieving a perfect congruent partition into N N pieces. I know no proof, no counter example. One can consider replacing 'congruent' with 'similar' in the above question. Some more related thoughts are in . References: A bit added on March 15th, 2024: Are there convex polygonal regions with even number of edges that has this property of being perfect congruent partitionable only into 2 non-convex pieces? mg.metric-geometry discrete-geometry plane-geometry tiling Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Jun 16, 2024 at 16:53 Martin Sleziak 4,783 4 4 gold badges 38 38 silver badges 42 42 bronze badges asked Aug 28, 2020 at 8:36 Nandakumar RNandakumar R 7,109 3 3 gold badges 9 9 silver badges 23 23 bronze badges 2 3 If you do not demand the regions to be simple polygons and congruence includes rotations, the counterexample is simple:take the regular hexagon and build something on its sides in the pattern aabbcc. Then you can split it into two congruent pieces by connecting the center to the "first" a,b,c and then to the "second" but no convex partition (i.e, splitting with a single line) into two congruent pieces will be possible. If you mean something more restricted, just say what exactly.fedja –fedja 2020-08-30 13:00:09 +00:00 Commented Aug 30, 2020 at 13:00 Thanks. The partition is to be into simple polygonal regions. Made a couple of edits to the question.Nandakumar R –Nandakumar R 2020-08-31 17:18:18 +00:00 Commented Aug 31, 2020 at 17:18 Add a comment| 1 Answer 1 Sorted by: Reset to default This answer is useful 20 Save this answer. Show activity on this post. This picture looks like a counterexample with N=2 N=2 and R R a convex pentagon: This should work more generally starting from an n×(n+1)n×(n+1) rectangles for any integer n>1 n>1, removing two congruent right triangles that are not isosceles (the picture shows n=3 n=3 with 2:5 2:5 triangles). Replacing the heavy lines with more complicated polygonal convex paths yields convex polygons with more sides. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Sep 1, 2020 at 1:13 Noam D. ElkiesNoam D. Elkies 81.2k 15 15 gold badges 288 288 silver badges 381 381 bronze badges 2 Beautiful! Nice to use a pentagon that cannot be partitioned into two congruent convex pieces, as that would require a chord to create identical angles.Joseph O'Rourke –Joseph O'Rourke 2020-09-01 15:19:43 +00:00 Commented Sep 1, 2020 at 15:19 2 Really elegant example; thanks!! Let me add a small extension to the question, basically to bound this property: to find that convex shape with such a perfect congruent partition into 2 non-convex pieces that 'wastes' the highest fraction of its area when given the best partition into 2 convex pieces.Nandakumar R –Nandakumar R 2020-09-01 17:11:08 +00:00 Commented Sep 1, 2020 at 17:11 Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions mg.metric-geometry discrete-geometry plane-geometry tiling See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked 3Triangles that can be cut into mutually congruent and non-convex polygons 5Are there convex polyhedrons that can be cut into mutually congruent connected pieces only if pieces are non-convex? 0On partitioning convex polygons into mutually congruent spiral polygons 1An algorithm to decide whether a convex polygon can be cut into 2 mutually congruent pieces Related 2On congruent partitions of planar regions 1On convex planar regions that can be cut into only a specified number of mutually congruent and connected pieces 3Triangles that can be cut into mutually congruent and non-convex polygons Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. 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https://www.sciencedirect.com/science/article/pii/S0021925820302404
The pneumococcal σX activator, ComW, is a DNA-binding protein critical for natural transformation - ScienceDirect Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Introduction Results Discussion Materials and methods Author contributions Acknowledgments Supplementary Material References Show full outline Cited by (10) Figures (8) Show 2 more figures Extras (1) Document Journal of Biological Chemistry ------------------------------- Volume 294, Issue 29, 19 July 2019, Pages 11101-11118 Gene Regulation The pneumococcal σ X activator, ComW, is a DNA-binding protein critical for natural transformation Pneumococcal ComW binds to DNA Author links open overlay panel Nicole L.Inniss‡, Gerd Prehna§, Donald A.Morrison‡ Show more Outline Add to Mendeley Share Cite rights and content Under a Creative Commons license Open access Natural genetic transformation via horizontal gene transfer enables rapid adaptation to dynamic environments and contributes to both antibiotic resistance and vaccine evasion among bacterial populations. In Streptococcus pneumoniae (pneumococcus), transformation occurs when cells enter competence, a transient state in which cells express the competence master regulator, SigX (σ X), an alternative σ factor (σ), and a competence co-regulator, ComW. Together, ComW and σ X facilitate expression of the genes required for DNA uptake and genetic recombination. SigX activity depends on ComW, as Δ comW cells transcribe late genes and transform at levels 10- and 10,000-fold below that of WT cells, respectively. Previous findings suggest that ComW functions during assembly of the RNA polymerase-σ Xholoenzyme to help promote transcription from σ X-targeted promoters. However, it remains unknown how ComW facilitates holoenzyme assembly. As ComW seems to be unique to Gram-positive cocci and has no sequence similarity with known transcriptional activators, here we used Rosetta to generate an ab initio model of pneumococcal ComW's 3D-structure. Using this model as a basis for further biochemical, biophysical, and genetic investigations into the molecular features important for its function, we report that ComW is a predicted globular protein and that it interacts with DNA, independently of DNA sequence. We also identified conserved motifs in ComW and show that key residues in these motifs contribute to DNA binding. Lastly, we provide evidence that ComW's DNA-binding activity is important for transformation in pneumococcus. Our findings begin to fill the gaps in understanding how ComW regulates σ X activity during bacterial natural transformation. Previous article in issue Next article in issue Streptococcus microbiology RNA polymerase DNA transformation protein structure bacterial transcription competence ComW pneumococcus sigma factor Introduction Bacterial natural genetic transformation is the uptake and incorporation of exogenous DNA into a cell's genome. Transformation was discovered in Streptococcus pneumoniae (pneumococcus) (1, 2), and subsequently the genes required for transformation have been found in the genomes of streptococci from all species groups (3). As a result, streptococci have highly malleable genomes, making them well equipped for adaptation, as seen with frequent capsular switching, and the rapid spread of genes that mediate antibiotic resistance (4). Natural transformation has increased the diversity of the pneumococcal genome, directly contributing to the evolution of mutli-drug resistant pneumococcal strains (4). As over 82 bacterial species are naturally transformable (5), and many of these are important human pathogens, a deep understanding of this horizontal gene transfer mechanism is important for continued progress in the fight against drug resistant pathogens. Streptococci primed for transformation are described as competent. Competence is a transient state marked by a shift in both transcriptomic and proteomic profiles (67, 89). Competence entry is controlled by production of the alternative σ factor (σ),3 SigX (σ X), a member of the σ 70 family of proteins (10, 1112). σ factors transiently associate with core RNA polymerase (E) to direct the holoenzyme (Eσ) to specific promoters to initiate transcription (13, 14). Like all bacteria, streptococci contain a principal σ factor, SigA (σ A), responsible for most gene transcription. Many bacteria have multiple alternative σ factors, mediating responses to diverse challenges. However, streptococci have a single alternative σ factor, σ X. Interestingly, some streptococcal species contain multiple copies of sigX (also known as comX) (3), and sigX expression is strictly linked to competence development. Streptococci utilize two tightly regulated quorum-sensing systems to coordinate σ X-mediated competence and other group behaviors (1516, 1718). Activation of the ComRS or the ComCDE system coordinates uniform sigX expression. Species of the bovis, pyogenic, salivarius, and suis groups utilize the ComRS system to drive σ X production (15, 16). In this system, the pro-peptide ComS is cleaved to its mature form, XIP (Sig X I nducing P eptide). XIP (ComS) is actively exported and reimported for binding to ComR, a member of the Rgg, Rap, NprR, PlcR, and PrgX (RRNPP) family of transcriptional regulators (19, 20). The ComR-XIP complex (or ComRS) interacts with DNA as a dimer (21, 22), and binds to a conserved inverted repeat upstream of sigX, termed the ComR-box (7, 23), to activate transcription of the competence regulon. In contrast, species of the angionsus and mitis groups use the ComCDE pathway (17), an auto-regulated Two-component Signal Transduction system (TCST) that responds to the C ompetence S timulating P eptide (CSP) (24). Interestingly, Streptococcus mutans contains both pathways (15, 25), adding additional layers of complexity to its competence regulation. Therefore, although the competence master regulator, σ X, is conserved, streptococci have evolved multiple molecular mechanisms to regulate its expression and consequently, competence. Much of our understanding of streptococcal transformation comes from initial work done with the ComCDE system in pneumococcus. During exponential growth, the Eσ A basally transcribes comC leading to production of the pro-peptide, ComC (24). ComC is simultaneously cleaved and exported as CSP by the ABC-transporter, ComAB (comAB) (26, 2728). CSP is sensed by the histidine-kinase receptor, ComD, resulting in auto-phosphorylation and activation of its cognate response regulator, ComE, via a phospho-relay event (29, 3031). Activated ComE promotes transcription of comCDE and other early competence genes (32). ComE-mediated transcription triggers robust competence among the cell population in an auto-positive-feedback regulatory loop. This response culminates in ComE dependent production of σ X (18, 32). During competence, Eσ X transcribes from the combox promoter, directly linking ComCDE quorum sensing to the expression of the transformation regulon (18). ComE also activates expression of a competence co-regulator, comW, encoding ComW, a 9.5-kDa protein of unknown function (33). ComW homologs are only found in eleven species of the anginosus and mitis groups. Like σ X, ComW is transient and tightly controlled by ComCDE, suggesting that ComW is unique to ComCDE competence induction. Although σ X is the master regulator and sigX is expressed independently of ComW (33, 34), σ X activity seems to be dependent on the presence of ComW (35). SigX's requirement for ComW is supported by observations that mutants lacking ComW (Δ comW) transcribe late genes and transform at levels 10- and 10,000-fold below that of WT cells, respectively (36). Furthermore, Δ comW cells have decreased σ X levels (35), and ComW and σ X weakly interact (37), suggesting that they physically function together in the cell. As multiple σ factor subunits are present in a cell simultaneously, bacteria have evolved many precise and rapid mechanisms to regulate σ activity. The most common regulatory mechanism of alternative σ activity is sequestration by an anti-σ, and subsequent release of σ upon sensing of the appropriate cellular or environmental cues. Release from sequestration allows alternative σ factors to interact with E resulting in specific Eσ formation (38, 39). Less common methods of σ factor activation include small protein activators, like the Gram-negative specific Crl (40) (an Eσ S assembly factor) and two-part σ factors, as seen in some bacteriophage systems (4142, 43, 4445) and in Bacillus subtilis (46). In an attempt to pinpoint which of these mechanisms may be applicable to ComW-mediated σ X activation, a suppressor screen was conducted in Δ comW pneumococcal background. Remarkably, only mutations in σ A partially restored late gene expression and transformation efficiency (36, 37). These σ A suppressor mutations were restricted to σ factor regions 2 and 4 (σ 2, σ 4) (36, 37), domains involved in E binding and promoter recognition (47). We interpret these results as indicators that ComW functions during the swap from reliance on Eσ A- mediated transcription to reliance on Eσ X- mediated transcription during competence and not as a component of a sequestration system. SigX co-purifies with E from competent pneumococci; but ComW was not isolated with the Eσ X complex (48). Thus, it is unknown if binding to E is required for ComW function, or whether ComW specifically acts at combox promoters. Purified Eσ X can direct transcription from combox promoters in vitro without ComW (12). However, these assays were not done in competition with σ A, leaving gaps in our understanding of how ComW promotes Eσ X formation over that of Eσ A at the onset of competence. As these previous attempts to define how ComW works have not involved direct biochemical or biophysical examination of the protein, our current understanding of ComW's role during competence is based exclusively on indirect, genetic evidence. To directly explore the biochemical, biophysical, and genetic properties of ComW to help decipher its function as an Eσ X assembly factor, we created a structural model of ComW using Rosetta (49). Using this model we identified ComW as structurally homologous to DNA binding proteins and show that ComW directly binds to DNA, independent of DNA sequence. Additionally, we show that specific residues on ComW's conserved molecular surface are involved in DNA-binding activity and link this activity to efficient transformation in pneumococcus. Results ComW is a protein unique to Gram-positive cocci ComW was initially identified as a CSP responsive competence regulator in multiple pneumococcal strains (18, 33, 50). An alignment of 19 ComW proteins from different pneumococcal strains showed that most alleles share a high percentage of residues that are identical to the R6 sequence. However, pneumococcal ComW alleles can vary by as much as 26% (Fig. 1A). This suggests that ComW is conserved within a species and likely plays an important role for the pneumococcal competence response. Furthermore, comparisons of sequences within and across species suggest that there is intra- and inter-species conservation of ComW. 1. Download: Download high-res image (569KB) 2. Download: Download full-size image Figure 1. Multi-sequence alignment of ComW.A, alignment of 19 ComW proteins from different pneumococcal strains. A BLAST with the ComW sequence from an R6 strain (NP_357614.1) returned 63 different ComW alleles. Eighteen alleles, spanning the full range of diversity for the alleles identified, were selected at random, for alignment to R6 ComW. (National Center for Biotechnology Information accession number): R6 (NP_357614.1), SMRU2065 (WP_050233459.1), 27 (ETE01301.1), D39 (ABJ54252.1), 2082239 (EJG75956.1), 430772 (WP_0543900.1), BS70 (EDK62216.1), BS74 (EDK68700.1), CDC108700 (EDT90831.1), 2849(), BS72 (EDK80769), PNI0009 (ELU82677.1), R34 (CTL40368.1), 08B03195(WP_050216603.1), SMRU964 (WP_050235807.1), SMRU2689 (WP_050216449.1), N (WP_050209129.1), SMRU2014 (WP_05024327.1), SMRU824 (WP_050253536.1). Note: Few pneumococcal sequences contain an alternative start codon (TTG) six bases upstream of the ATG start (not shown). As most comW sequences in pneumococci and other streptococci lack this TTG start codon, and it is in close proximity to the ribosome-binding site, the alternative start codon is not considered as part of the ComW sequence. B, alignment of ComW orthologs from S. pneumoniae (Sp, NP_357614.1), S. pseudopneumoniae (Sps, AEL09536.1), S. mitis (Sm, KEQ48646.1), Peptoniphiluslacrimalis (DNF00528), S. sinensis (Sn, WP_037617413.1) Streptococcus cristatus (Sc, EGU68430.1), S. infantis (Si, EFO53736.1), S. anginosus (Sa, EJP26452.1), S. oralis (So, WP_000939510.1), S. dentisani (Sd, WP_038804352.1). For A and B, black text, residues identical to R6; red text, resides that differ from R6; asterisks, identical residues; colons, conserved residues; periods, semi-conserved residues. ComW orthologs from S. anginosus, S. mitis, S. oralis, and S. pseudopneumoniae were moved to pneumococcus, and tested for transformation efficiency. C, Espript3 Alignment of ComW orthologs from streptococci showing helical placement and solvent exposed residues based on a ComW model. Red shading, 100% conserved residues; red letters, similar within a group, residues framed in blue, similar across groups. Relative surface accessibility (acc) of each residue is indicated: dark blue box for accessible, cyan box for intermediate, and white box for buried. BLAST searches with the ComW sequence from the R6 pneumococcal strain revealed ComW orthologs conserved in at least ten other streptococcal species. These species belong to the anginosus (S. anginosus, S. cristatus, S. oligofermentans, and S. sinensis) and mitis groups (S. dentisani, S. infantis, S. mitis, S. oralis, S. pseudopneumoniae, and S. tigurinus). Initial examination of such ComW orthologs showed that many retain a high percentage of residues that are identical to the pneumococcal protein, like ComW from S. mitis (ComW Sm, 74%) and S. pseudopneumoniae (ComW Sps, 86%), whereas others have as little as 41% identity, like ComW from S. dentisani (Fig. 1B). Like pneumococcus, S. dentisani, S. mitis, and S. pseudopneumoniae are members of the mitis group, suggesting that ComW alleles vary across closely related species. Intra-species variation of ComW is common within S. mitis (Fig. S1). S. mitis ComW proteins that are most similar to pneumococcal ComW can share 88% residue identity. However, some S. mitis ComW proteins share only 40% sequence identity with the ComW of the R6 pneumococal strain. These relationships suggest that ComW is conserved across many members of the anginosus and mitis groups, but also highlights the divergence of ComW sequences, both at the inter- and intra-species level. Analysis of ComW sequences revealed four highly conserved regions. In the pneumococcus strain R6, these regions are 17 EEEY 20, 29 DWE 31, 38 LIYYLVR 44, and 56 YHYRVAYRLY 65. Although ComW sequences exhibit variation, specific aromatic and charged residues within these regions are absolutely conserved in all available orthologs. These four motifs are: 17 ExEY 20, 29 xWE 31, 38 LxYYLxR 44, and 56 YH(Y/F)RxxYRxY 65 (Fig. 1B). In addition, many of the residues in these motifs are predicted to be solvent exposed, based on a ComW model (Fig. 1C, Fig. 2). ComW sequences also contain shorter variable regions. Of particular interest, the R6 C-terminal residues, 73 RGFISC 78, are varied or missing from some orthologs (Fig. 1B). These patterns suggest that residues in the conserved motifs are key for ComW function, whereas the C terminus may be dispensable in some species. 1. Download: Download high-res image (323KB) 2. Download: Download full-size image Figure 2. An ab initio Rosetta model of pneumococcal ComW.A, a ribbon view of the ComW model, colored according to secondary structure (α-helices are coral, loops and disordered C-terminal residues (73 RGFISC 78) are light gray). B, residue conservation was determined using Consurf and displayed on a molecular surface model using Chimera (Petersen, 2004). Residues of the 17 EEEY 20, 38 LIYYLVR 44, and 56 YHYRVAYRLY 65 motifs, and conserved, solvent exposed tyrosines (Y) are circled. Residues targeted for in vivomutagenesis in this study are written in red. The nonconserved C-terminal (73 RGFISC 78) residues are shaded by gray circles. C, predicted surface electro statics as determined using APBS (pH 7.0, parse) and drawn in Chimera; red, negatively charged; blue, positively charged, (−10 to 10 kT/e). Conserved motifs, tyrosines, and C-terminal residues are labeled as in B. A BLAST search also identified an ortholog of ComW in the Gram-positive, anaerobic coccus (GPAC), Peptoniphiluslacrimalis (51). Only 47% of P. lacrimalis' ComW residues are identical to those of the pneumococal R6 strain, and the conserved motifs are retained. The full length ComW sequence was not found in any Gram-negative organisms. Thus, ComW seems to be unique to Gram-positive bacteria. ComW is critical for pneumococcal competence, but its function is unknown, and it lacks close relatives outside of Gram-positive cocci. This rarity impeded production of a homology model, for example by Phyre2D (52), and suggests that ComW may adopt a novel fold that is important to its function. A structural model of ComW We pursued a structural model to probe ComW's biochemical and biophysical properties. Attempts to crystallize ComW were unsuccessful, and NMR studies proved cumbersome due to the buffer conditions required for long term stability of the protein above 4 °C (50 m mTris, pH 8.0, 500 m m NaCl, 10% glycerol, 1 m m EDTA, 1 m m βME) (Fig. S2). As ComW has no close relatives with known structures for the creation of a homology model (Phyre2), we used Robetta (53) and Rosetta (49) to calculate an ab initio ComW model. Initial models of pneumococcal ComW were built using the Robetta server (53). Robetta produced five models (models not shown). Each model was of a globular protein, composed of 3–4 α-helices connected by loops, with a disordered C-terminal tail (residues 73 RGFISC 78). Interestingly, Robetta predicted similar models for ComW from S. pseudopneumoniae (ComW Sps, 86% residue identity to ComW from the pneumococcus R6 strain) and ComW from S. anginosus (ComW Sa, 43% residue identity to ComW from pneumococcus R6 strain) (models not shown). These initial predictions suggested that, despite primary sequence variability, all ComW proteins share a similar fold. To directly control model building parameters, as described in the methods, we also used a local installation of Rosetta (49) to model pneumococcal ComW. The Rosetta calculation produced 80,000 ComW models that were similar to those created by the Robetta server. To find a suitable model for ComW, we narrowed the model pool to the 8,000 lowest energy structures, and clustered the models based on similarity. A summary of five Rosetta low energy clusters is shown in Fig. S3, A–E. Overall the models are globally similar 4-helix bundles with some variation in helical packing. The lowest energy models from each cluster are superimposed in Fig. S3 F, and are similar in helical content but differ in their helical packing. This overall structural similarity suggests that the Rosetta calculation produced many models that converge on a similar fold. Models in Clusters 0 and 1 were nearly identical in energy score (Fig. S3), making it difficult to distinguish which model is most representative of ComW's native state. We used the DALI server (54) to determine which cluster likely contained models most similar to folds of known function. As detailed later, the lowest energy model from Cluster 1 yielded high quality alignments with proteins or domains of similar size, as determined by DALI Z-scores, indicating it was closer to known protein folds. This low energy structure was chosen as the representative ComW model, and is shown in Fig. 2A. The Rosetta model of S. pneumoniae's ComW is a globular protein, composed of four tightly packed α-helices that are connected by one short linker and two loops, with a disordered C-terminal tail (residues 73–78), placing the N and C termini at opposite sides of the molecule (Fig. 2A). Surface representations (Fig. 2, B and C) show that the tightly packed helices form a solid core with multiple exterior grooves and/or pockets, in suitable positions to act as binding sites for other biomolecules. To visualize surface amino acid conservation the Consurf server (55) was used. From the conservation map, the model predicts that one face of ComW has a large area of highly conserved residues (Fig. 2B, right). This face is composed of elements from two conserved motifs; residues L38, Y40, Y41, L42, V43, and R44 (38 LIYYLVR 44) in helix α3, and Y56, H57, R59, Y62, and R63, (56 YHYRVAYRLY 65) in helix α4 (Fig. 1B and 2B). Residues Y40, Y41, and R44 are conserved across species and are specifically predicted to be solvent exposed. Residues Y56, H57, R59, Y62, and R63 are also conserved across species and are also predicted to be solvent exposed on this face or the bottom of ComW. The rest of the residues from the 38 LIYYLVR 44 and 56 YHYRVAYRLY 65 motifs are mostly buried in the ComW's hydrophobic core. On the opposite face of pneumococal ComW, there is no contiguous stretch of conserved residues (Fig. 2B, left). However, the strictly conserved E17 residue of the 17 EEEY 20 motif in helix α1 is predicted to be partially solvent exposed and the center of a deep pocket. Residues E18 and E19 are fully exposed, Y20 is partially solvent exposed, and all three of these amino acids extend to ComW's bottom. To calculate the electrostatic potential the Adaptive Poisson Boltzmann Solver (APBS) (56) was used (Fig. 2C). Interestingly, ComW's opposing faces have opposite electrostatic potential, as calculated by APBS (Fig. 1C). The nonconserved face is highly electronegative, especially the pocket centered on E17 (Fig. 1C, left), whereas the conserved face is electropositive (Fig. 1C, right). Based on these structural features, we hypothesized that these molecular surfaces are binding sites for a protein or other biomolecule and that specific residues in the conserved motifs are integral to such interactions. ComWΔ6 oligomerizes in solution Although we have yet to obtain an X-ray crystal or NMR structure of ComW, a truncated version lacking C-terminal residues 73 RGFISC 78 (ComWΔ6, 12.4 kDa with V5H6 tag) was purified via affinity chromatography (Fig. S4) for biochemical and biophysical characterization. A circular dichroism (CD) spectrum indicated a predominately α-helical structure in agreement with the Rosetta model (Fig. 3). 1. Download: Download high-res image (88KB) 2. Download: Download full-size image Figure 3. CD profiles of ComWΔ6 variants in solution. ComWΔ6 variant proteins were diluted to 0.02 mg/ml for secondary structure analysis. Deconvolution was performed using Dichroweb. Green, ComWΔ6; blue, E18A; orange, 38 LxYYLxR 44 mutants. Size exclusion chromatography (SEC) experiments suggested that ComWΔ6 oligomerizes in solution, as the pure protein eluted from the gel filtration column at a molecular weight of 25.5 kDa, indicative of a dimer (Fig. 4A). Analytical ultra-centrifugation (AUC) sedimentation velocity experiments were performed to verify the oligomeric state of ComWΔ6 (Fig. 4B). The ability of ComWΔ6 to oligomerize at increasing protein concentrations (OD 280 of 0.5 (0.31 mg/ml), 1.0 (0.61 mg/ml), and 1.5 (0.92 mg/ml)) was examined. Analysis of sedimentation data revealed one primary peak for each concentration analyzed. At concentrations of 0.31 mg/ml and 0.61 mg/ml, over 60% of ComWΔ6 sedimented at 2.04S (MW ∼16.6 kDa), and 2.12S (MW ∼17.0 kDa), respectively. Although some higher molecular weight aggregates were observed, the above sedimentation values are largely indicative of a monomeric state. At a concentration of 0.92 mg/ml, 95% of ComWΔ6 sedimented at 2.693S, with a calculated molecular weight of 25.6 kDa in agreement with SEC experiments. This shows that ComWΔ6 has a propensity to form dimers in solution, and that dimerization is concentration dependent. 1. Download: Download high-res image (84KB) 2. Download: Download full-size image Figure 4. Quaternary structure of ComWΔ6 variants.A,size exclusion chromatography elution profiles of ComWΔ6 variants. The green curve is ComWΔ6, and the orange curves are the L42A and R44A mutants. Top table shows elution volumes and experimental molecular weights of four protein standards. Protein standards were used to calculate the experimental molecular weights of ComWΔ6 variants. Bottom table shows the theoretical and experimental molecular weights, plus the elution fraction (ml) of ComWΔ6 variants. B,sedimentation velocity profiles of ComWΔ6 at varied concentrations. Analytical ultracentrifugation was performed at Abs = 0.5 (gray peak, 0.31 mg/ml, 2.04S, 16.6 kDa), 1.0 (black peak, 0.61 mg/ml, 2.12S, 17.0 kDa), and 1.5 (green peak, 0.92 mg/ml, 2.7S, 25.6 kDa). Vbar (0.736), buffer density (d = 1.02g/L), and viscosity (Poise = 1.07 A10−2) were determined using SEDNTERP. SEDFIT was used for continuous distribution and Bayesian analysis of the data. The ComW model is similar to σ factor structures The ab initio ComW models produced by Rosetta were used to search the Protein Data bank (PDB) (www.rcsb.org) (57) for similar protein folds, and to help validate our Rosetta calculation. We submitted one model from each Rosetta-generated cluster (Fig. S3), to the D istance Ali gnment Matrix (DALI) (58) server. Searches with each model returned 600 hits as structurally similar proteins with known function. The combined search yielded 3,000 total hits. Some structural hits were repeated within and across model searches. To focus on proteins that may offer clues to ComW's biological function, DALI hits from all five searches were compared. Proteins that were less than 200 amino acids in length, and/or had DALI Z-scores above 4.0 were considered. Protein structures from the search displayed functions in protein binding, protein degradation, nucleic acid binding, and transcriptional regulation. Notably, searches with each model returned the Escherichia coli (E. coli) primary σ, σ 70 (Z-score = 5.5) (59), as structurally similar. In addition, a number of extra cytoplasmic (ECF) σ factors, including σ W of Bacillus subtilis (B. subtilis) (Z-score = 5.7) (60), σ E of E. coli (Z-score = 5.8) (61) (Fig. 5), and σ K of Mycobacterium tuberculosis (M. tuberculosis) (Z-score = 5.6) (62), were structurally similar to the ComW model. Interestingly, all superimpositions of ComW with these σ factors identified by DALI showed that ComW is structurally homologous to the σ 2 domain. This domain is known to directly interact with RNA polymerase, and function during −10 promoter element recognition and dsDNA melting (63, 64). Thus although ComW has no sequence homologs outside of Gram-positive cocci, it likely adopts a fold similar to some σ-factors that exist in many bacteria. Given that both σ X-mediated transcription and pneumococcal transformation are dependent on ComW (37), these data strongly suggest that ComW acts as a DNA binding protein at the onset of competence. 1. Download: Download high-res image (127KB) 2. Download: Download full-size image Figure 5. Two structural alignments of the ComW model and selected σ factors.A, Chimera (Petersen, 2004) was used to superimpose the ComW model (coral) on the structurally similar σ factor, σ E of E. coli. B, Chimera was used to superimpose the ComW model (coral) on the structurually similar σ factor, σ W of B. subtilis. For both panels, the PDB entry and aligned residues, polypeptide chain, Z-score (calculated by the DALI server, and considers matched residues and domain size), root-mean-square-deviation (rmsd), and functional σ region are given in each table, for each structure. ComWΔ6 binds to DNA, nonspecifically To test for DNA-binding activity, ComWΔ6 was used in electrophoretic mobility shift assays (EMSA). Increasing amounts of ComWΔ6 were incubated with a fluorescently labeled probe containing the σ X competence specific promoter, known as the combox, (65) (Fig. 6A). An increase in the amount of labeled DNA bound by protein was observed as increasing amounts of ComWΔ6 were added to EMSA reactions (Fig. 6C, left). Additionally, a control with cytochrome C showed that the results were likely not artificial protein-DNA interaction simply from the use of concentrated purified protein (Fig. S5). Thus ComWΔ6 can bind to DNA in the absence of any other protein partners. 1. Download: Download high-res image (127KB) 2. Download: Download full-size image Figure 6. Electrophoretic mobility shift of two DNA probes with ComWΔ6.A, schematic of DNA probes used in EMSA. Top, a schematic of pneumococcus' combox promoter region, positioned upstream of the late competence gene, ssbB. Bases highlighted in blue are the conserved −10−25, and −35 promoter elements. Red bases are required for transcription from the combox promoter according to Campbell et al. (1998). Bottom, a schematic of S. mutan's ComR-box region, upstream of the sigX gene. Pink residues are the conserved imperfect, inverted repeat recognized by the ComRS complex. B, binding curves of DNA probe shifts seen with increasing amounts of ComWΔ6 from C. Curves represent averages from three biological replicates. C, ComWΔ6 titration, up to 32μM, with 20 n m labeled probe, combox (left) or ComR-box (right). Both probes are 589 bp in length. Representative gels from three biological replicates are shown. To determine if the ComWΔ6-DNA interaction was specific for the combox promoter, a labeled probe containing the region upstream of sigX from Streptococcus mutans (Fig. 6B) was used in DNA binding assays. In S. mutans, ComRS, a member of the Rgg-like family of transcriptional regulators, controls expression of sigX from the ComR-box promoter, a 20-bp imperfect, inverted repeat (7, 15, 23). This promoter region differs greatly from the combox promoter that is targeted for transcription by σ X (Fig. 6A). As ComW has only been identified in streptococcal species that utilize the ComCDE competence activation pathway, we predicted that ComW would not bind to ComR-box containing DNA. However, an increase in the amount of shifted S. mutans probe was observed when increasing amounts of ComWΔ6 were added to the reaction (Fig. 6C, right). A comparison of binding curves of these interactions (Fig. 6B) suggested that ComWΔ6 binds to each probe with similar kinetics, and therefore binding is not dependent on DNA sequence. ComWΔ6 was also able to shift a plasmid derived DNA probe in EMSA (not shown), further supporting that ComWΔ6 can interact with DNA independently of sequence. In addition, ComWΔ6 interacted with a single stranded DNA (ssDNA) probe only at >64μM of protein (not shown), suggesting that ComWΔ6 may prefer dsDNA targets. Mutations to the 38 LIYYLVR 44 motif of ComW disrupt DNA binding Structural alignments of the ComW model with alternative σ factors showed that ComW likely adopts a similar fold to these transcriptional regulators (Fig. 5). Close examination of structural alignments between the ComW model and the crystal structure of σ 2 of E. coli's σ E (61) shows that many of the residues that constitute ComW's conserved surface align with residues that are important for σ E-DNA contact during promoter recognition and melting. To investigate if residues in this region of ComW played a role in DNA binding, we purified mutants in the 38 LIYYLVR 44 motif, and one mutant in the 17 EEEY 20 motif located on the opposite electronegative face. Only ComWΔ6 E18A, ComWΔ6 Y40A, ComWΔ6 L42A, and ComWΔ6 R44A proved soluble and stable in vitro, as demonstrated by CD (Fig. 3). Their spectra indicated predominately α-helical structures, again in agreement with the Rosetta model. Furthermore, these data indicate that the point mutations do not disrupt the structure relative to ComWΔ6. Thus, these variants were used in our biochemical assay. Each soluble ComWΔ6 variant was tested for DNA binding by EMSA with combox and ComR-box DNA. As expected, the nonconserved E18A mutant was able to interact with both DNA probes, indicating that this residue was dispensable for DNA binding (Fig. 7B, left and middle). Comparisons of DNA binding curves also suggested ComWΔ6 E18A mutant binds to different probes with similar kinetics; therefore binding was independent of sequence (Fig. 7B, right). ComWΔ6 Y40A also interacted with both DNA probes, indicating that this residue was dispensable for DNA binding (Fig. 7C, left and middle). Again, DNA binding curves showed similar binding kinetics with pneumococcal and S. mutans probes (Fig. 7C, right), further supporting that DNA binding is not dependent on DNA sequence. At 4 μM protein, ComWΔ6 E18A and ComWΔ6 Y40A mutants did not differ in their ability to bind to the DNA probes when compared with ComWΔ6 (Fig. 7A, left). Interestingly, unlike ComWΔ6, ComWΔ6 E18A and ComWΔ6 Y40A point-mutants showed increased total DNA shift at 32 μM, irrespective of DNA sequence, and appeared to shift the DNA to a higher oligomeric form (Fig. 7, A, right and B and C). Mutation of residues E18 or Y40 to alanine may increase total binding of ComWΔ6 to DNA. 1. Download: Download high-res image (185KB) 2. Download: Download full-size image Figure 7. Electrophoretic mobility shifts of two DNA probes with ComWΔ6 variants.A, comparison of DNA probe shifted by ComWΔ6 variants at 4μM (left) and 32μM (right). For all samples, n = 3 biological replicates. C, pneumo combox probe; R, S. mutans ComR-box probe. Asterisks () mark statistically significant differences between ComWΔ6 and mutants bound to probe (, p< 0.05; , p< 0.01). B–E, representative gels of mutant ComWΔ6 titrations, up to 32μM, with 20 n m labeled probe, combox (left) or ComR-box (middle), and binding curves (right). Both probes are 589 bp in length. The average curves from three biological replicates are shown. In contrast, ComWΔ6 L42A and ComWΔ6 R44A showed a decrease in ability to interact with combox or ComR-box DNA, when compared with ComWΔ6. ComWΔ6 L42A binds to DNA at ≥4.0μM, compared with binding at 1.0 μM as seen with ComWΔ6 (Fig. 7, A and D, left and middle, and Fig. 6C). However, at 32 μM, ComWΔ6 L42A appeared to bind the DNA probes at levels that were not significantly different than ComWΔ6 (Fig. 7A, right). In addition, a comparison of DNA binding curves with different probes, suggested that ComWΔ6 L42A retains the ability to bind to DNA independently of sequence (Fig. 7D, right). ComWΔ6 R44A showed the most dramatic decrease in DNA binding. At 4 μM, ComWΔ6 R44A did not bind to either DNA probe (Fig. 7E, left and middle, Fig. 7A, left). More than 8 μM of protein was required to shift small amounts of the probe in EMSA gels (Fig. 7E, left and middle). At 32 μM of protein, ComWΔ6 R44A shifted only 11–12% of either DNA probe (Fig. 7A, right), significantly less DNA than that shifted by ComWΔ6. This dramatic decrease in binding activity is evident in DNA binding curves. However, the DNA binding curves did show that there was no significant difference in binding to pneumococcal and S. mutans probes (Fig. 7E, right) by ComWΔ6 R44A. These results demonstrate that residues ComWΔ6 L42 and ComWΔ6 R44 participate in the ComWΔ6-DNA interaction. Thus, the 38 LIYYLVR 44 motif, present on the positively charged conserved face of ComW, is important for efficient DNA binding. Residues that participate in ComWΔ6 DNA binding are not required for oligomerization As ComWΔ6 oligomerizes (Fig. 4), which may contribute to DNA binding, we used SEC to determine the integrity of quaternary structure in the ComWΔ6 L42A and ComWΔ6 R44A mutants. Like ComWΔ6 (25.5 kDa), the ComWΔ6 L42A and ComWΔ6 R44A mutants eluted as dimers, with molecular weights of 19.4 and 24.4 kDa, respectively (Fig. 4). This demonstrated that mutations in the 38 LIYYLVR 44 motif of pneumococcal ComW did not disrupt oligomerization. Together, the SEC and EMSA results show that ComW's 38 LIYYLVR 44 motif on the conserved, electropositive face is unlikely to participate in protein oligomerization, but instead is part of a direct binding site for DNA. ComW mutations alter protein levels and pneumococcal transformation efficiencies To determine levels of ComW production and transformation efficiency in comW mutant strains, we used Western blots and a pneumococcal natural transformation assay. A pneumococcal-specific antibody detected WT ComW as a 9.5-kDa band on blots, and WT cells transformed with an efficiency of 66%. In contrast, Δ comW mutants produced no ComW and transformed at levels 10,000-fold below that of WT cells (Fig. 8, A and C, and Fig. S6 B). We attribute the loss of transformability to complete loss of ComW protein in this strain. Following biochemical characterization of the ComWΔ6 variant, we also determined how removal of the disordered C-terminal domain altered ComW production and function in pneumococcus. ComWΔ6 was produced at levels that were 1/3 that of WT cells and comW Δ 6 mutants transformed at 31% efficiency (Fig. 8, A and C, and Fig. S6 B). Thus comW Δ 6-expressing pneumococci exhibited a decrease in transformability that appears to be dependent on the level of ComWΔ6 production. Thus, the C-terminal residues, 73 RGFISC 78, which are not conserved across streptococci (Fig. 1D), are important for ComW stability and full function in vivo, but are dispensable for DNA binding in vitro. The specific functional importance of the disordered tail has yet to be determined. 1. Download: Download high-res image (114KB) 2. Download: Download full-size image Figure 8. Function and expression of ComW variants in pneumococcus.A, transformation efficiency in pneumococcal cells expressing ComW mutant proteins. Residues mutated in the 17 EEEY 20 motifs are blue, in the 38 LIYYLVR 44 motif are orange, and the C-terminal truncation mutation (Δ 73 RGFISC 78) is green. For WT and ΔComW strains, n = 9; for ComW variant strains, n = 3. B, transformation efficiency in pneuococcal ComW chimera strains. Replacement of pneumococcal ComW with that of S. pseudopneumoniae (Sps), S. mitis (Sm), S. oralis (So), or S. anginosus (Sa). WT and ΔComW strains, n = 9 biological repliates; for ComW ortholog strains, n = 3 biological replicates. For both panels, after CSP induction of transformation and overnight growth on THY agar plates, supplemented with 10 μg/ml novobiocin, cells were counted and the #transformants/CFU determined. Error bars are standard deviation (S.D.); p ≤ 0.05, an asterisk () marks statistically significant differences in transformation efficiency compared with WT cells. C, representative images Western blotting detection of ComW variants in pneumococcal cells. 1 × 10 8 cells were added to each well. The RNA polymeraseβ subunit (∼137 kDa) was used as an endogenous loading control and to normalize ComW signal from each sample. We determined if changes to the 17 EEE 19 motif altered ComW levels and transformability of pneumococci. This motif is present on the electro-negative, nonconserved surface of ComW and does not appear to be required for DNA binding (Fig. 7, A and B). At 17 min post CSP induction, mutants ComW E17A, ComW E18A, or ComW E19A were produced at significantly lower levels than ComW (Fig. 8C, Fig. S6 B). Interestingly, the comW E17A and comW E19A-expressing mutants transformed at only 24 and 33%, respectively, but the comW E18A expressing mutant transformed at 73% efficiency (Fig. 8A). Pneumococci did not produce detectable levels of ComW 17AAA19, and comW 17AAA19-expressing mutants transformed at levels similar to that of Δ comW cells (Fig. 8, A and C, Fig. S6 B). Although decreases in ComW E17A, ComW E19A, and ComW 17AAA19 result in decreases in transformation efficiency, decreases in ComW E18A does not, and suggests that decreased protein production does not always result in a decrease in transformation efficiency. In addition, as these ComW variants are less stable than WT near peak competence, but mutation to this motif allows DNA binding, we hypothesize that the electro-negative surface of ComW serves an important functional role that may be separate from ComW DNA binding activity. We also examined how changes to the 38 LIYYLVR 44 motif altered ComW levels and pneumococcal transformability. This motif is present on ComW's electro-positive, conserved surface (Fig. 2) and is required for DNA binding (Fig. 7). ComW Y40A, ComW Y41A, ComW L42A, and ComW R44A were all detectable in pneumococci, but ComW Y41A was the only mutant produced at significantly lower levels than ComW, and ComW 40AA41 was not detected (Figs. 8C, Fig. S6 B). Interestingly, comW Y40A, comW Y41A, comW L42A, comW R44A, and comW Y40AA41 expressing cells all showed significant deceases in transformation efficiency at 52%, 10%, 16%, 27%, and 0% efficiencies, respectively (Fig. 8A). As residues L42 and R44 are each important for DNA binding activity (Fig. 7), and point mutation of either residue disrupts transformation without depletion of the ComW variant protein, we interpret this as indication that DNA binding activity, facilitated by the electro-positive surface of ComW, plays a key role during competence development. The ComW ortholog from S. anginosus does not complement pneumococcal ComW Although all streptococci share competence specific genes, including the master regulator, σ X, they differ in competence activation pathways (3). ComW is only produced in streptococci that activate competence via the ComCDE quorum-sensing pathway (10, 33, 66). Therefore it is possible that ComW is unique to only the ComCDE system, and that streptococci have evolved multiple mechanisms to regulate competence via direct regulation of σ X activity. We explored how natural variation in ComW orthologues affected their ability to function in competent pneumococci. Pneumococcal comW was replaced with that of S. anginosus (ComW Sa), S. mitis (ComW Sm), S. oralis (ComW So), or S. pseudopneumoniae (ComW Sps), and we examined the production of ComW orthologues and the corresponding transformation efficiency in the resulting chimeric strains. ComW Sm and ComW Sps are 73 and 78% identical to pneumococcal ComW respectively (with identical 17 EEEY 20, 38 LIYYLVR 44, and 56 YHYRVAYRLY 65 motifs, and have varied sequences of C-terminal residues, Fig. 1B). Both ComW Sm and ComW Sps were detected in Western blots from competent chimeras at 17 min post CSP induction, but levels of ComW Sps were significantly decreased compare with pneumococcal ComW, whereas ComW Sm levels were not (Fig. 8C and Fig. S6 B). The S. mitis and S. pseudopneumoniae chimeras transformed at 56% and 55%, respectively, and only the decrease in transformation with comW Sps was statistically significant (Fig. 8B). Although it is possible that differences in amino acid sequence contribute to the observed differences in protein levels, these data do demonstrate that the S. mitis and S. pseudopneumoniae orthologues are produced in pneumococcus and are functional in pneumococcus. Thus ComW functionality is retained across closely related species. ComW Sa shares only 43% identity with ComW (with 17 EAEY 20, 38 LIYYLIR 44, and 56 YHYRAAYRWY 65 motifs, and lacks the disordered C-terminal domain, Fig. 1D). Similarly, ComW So shares 40% identity with pneumococcal ComW (including 17 EQEY 20, 38 LLYYLIR 44, and 56 YHFRAAYRLY 65 motifs, and contains a disordered C-terminal domain, Fig. 1D). ComW Sa protein was not readily detected in Western blots and, in stark contrast to other orthologues, chimeras expressing comW Sa transformed at levels 10,000-fold below WT cells, a phenotype like Δ comW cells (Fig. 8, B and C). This suggests that ComW Sa is not stably produced in pneumococcal cells. In contrast ComW So was detectable, albeit at significantly decreased levels compared with WT, yet chimeras expressing comW So transformed at 41% efficiency (Fig. 8, B and C). This result is consistent with the fact that observed decreases in some ComW variants do not always predict equal decreases in transformability. It is possible that the specificity of the pneumococcal ComW antibody hinders detection of the more divergent ComW Sa and ComW So alleles. However, combined with their decreased transformation efficiencies, these data suggests that ComW Sa and ComW So are not stable and/or fully functional in the pneumococcal competence system. Therefore, although ComW proteins are similar across species, the transformation and Western blotting data suggest that there are species-specific determinants that are important for ComW stability and function. Discussion Since ComW's identification, there have been limited advances in our understanding of its function. Although a genetic link to the shift from σ A to σ X - dependent transcription has been established (36, 37), comW had not been probed for mutations that disrupt pneumococcal transformation. Furthermore, direct biochemical characterization of ComW was stalled by a lack of soluble protein. Bioinformatic tools, like Robetta, Rosetta, and the DALI server (49, 53, 58), have laid a foundation for deeper analysis of ComW function. The ab initio model of ComW described here depicts a globular protein, α-helical in structure, with a DNA-binding-like fold. Combined predictive analysis of ComW's surface characteristics using the APBS (56) and Consurf (55) servers identified two opposing faces of ComW: one conserved, electro-positively charged face, and one nonconserved, electro-negatively charged face. In vitro characterization of ComWΔ6 (12.4 kDa, Δ 73 RGFISC 78), a soluble variant with α-helical structure as predicted by the model (Fig. 3), shows that the protein dimerizes in solution (Fig. 4). Importantly, we report for the first time that ComWΔ6 binds to DNA independently of DNA sequence (Fig. 6), an interaction that depends on residues that constitute the conserved, electro-positively charged face (Fig. 7). Additionally, conserved residues on both faces of ComW are important for pneumococcal transformation (Fig. 8). Interestingly, point mutations to ComW's nonconserved, electro-negative face appear to destabilize the protein more than point mutations to ComW's conserved, electro-positive face. The presence of stable protein for poor DNA binding mutants ComW L42A and ComW R44A suggest that ComW-DNA interactions are important for transformation. Thus, we have begun to fill the gaps in understanding ComW's function as a regulator of σ X activity during natural transformation. Pneumococcal ComW and σ X are protein partners that promote transformation-specific gene transcription (33, 35, 3637). SigX is a member of the σ 70 family of proteins (10), most similar to Group 4 σ factors, the Extra-cytoplasmic Function (ECF) σ factor family. ECF σ factors are small, as they are composed of only the σ 2 and σ 4 domains essential for E binding, promoter recognition, and promoter melting (39). Our ComW model resembles the structure of the σ 2 domain of E. coli's primary σ factor, σ 70 (59), and the structures of some alternative σ factors, like B. subtilis' σ W (67), and E. coli's σ E (61, 68). In addition, ComWΔ6 interacts with DNA in EMSAs. This is an activity that has been observed with both Gram-negative and Gram-positive σ factors, including the structurally homologous σ J of M. tuberculosis (69), PG0162 of Porphyromonas gingivalis (70), σ B of Staphylococcus aureus (71), and σ E of Vibrio alginolyticus (72). In concert with the requirement of ComW for robust σ X-dependent transcription, we predict that ComW is an active member of the Eσ X complex during pneumococcal late-competence gene transcription. Amino acids that are important for σ factor-DNA contact have been identified in structures of multiple σ factors or σ factor domains in complex with DNA. Conserved aromatic and basic residues in the σ 2 domain act in promoter melting and nonsequence-specific DNA binding (63, 7374, 75, 76, 7778). The ComW model has one conserved surface that is composed of motifs 38 LIYYLVR 44 and 56 YHYRVAYRLY 65. Many of these aromatic and charged residues align with amino acids that are important for E. coli's σ E-DNA contact during promoter recognition and melting (61). Furthermore, we determined that mutation to two pneumococcal ComW residues, L42 and R44 on the conserved face, are important for nonspecific DNA binding, and that ComW L42A and ComW R44A are stably produced in pneumococcus but disrupt transformation efficiency. These results are in agreement with current understanding of the types of amino acids that σ factors utilize for DNA interaction, and support a direct role for ComW in transcription activation at σ X promoters. We expect that additional residues within these motifs aid in promoter recognition and DNA binding. Although E alone can bind to DNA and promote transcription (64, 79, 80), σ factors interact with E to specify transcriptional targets (13). SigX does not require ComW for interaction with E (unpublished data), and Eσ X can transcribe from combox promoters in vitro, in the absence of ComW (12). However these assays did not determine Eσ X promoter specificity in the presence of σ A or σ A dependent promoters. Hence it will be important to determine if ComW alters the affinity of σ X for E, and if/how ComW affects Eσ X transcription in these more competitive contexts. ComW's resemblance to σ 2 domains is peculiar because this domain also functions in DNA sequence-specific promoter recognition (63, 64). However, ComW interacts with DNA nonspecifically. Thus, it is possible that Eσ X promoter specificity is determined by σ X and that ComW mediates Eσ X-DNA interactions via binding to the DNA phosphate-backbone, or in another manner that does not depend on DNA sequence. Sigma factor activity is often controlled via direct interaction by a regulatory protein. Canonically, anti-σ factors, many of which are membrane proteins, sequester ECF σ factors, inhibiting Eσ formation. Specific cues activate regulated intramembrane proteolysis (RIP) cascades, leading to the release of the ECF (8182, 8384). An anti-σ X protein has not been identified (37), making it unlikely that ComW-mediated σ X regulation occurs via alleviation of a sequestration mechanism. Noncanonical σ factor control via interaction with a small protein has been observed in both Gram-negative and Gram-positive organisms. In Gram-negative organisms, like E. coli, Crl (∼15.6 kDa) binds to the stress response specific σ S and to E, to promote formation of the Eσ S complex (40, 85). In the absence of Crl, Eσ S fails to form stable complexes, and transcription from σ S promoters is decreased (85). As we have observed Eσ X formation in competent pneumococcal Δ comW cells (48), the structure of Crl (compose of an α+β fold, (86)) vastly differs from the ComW model, and no Crl-DNA interactions have been reported, it is unlikely that ComW shares an equivalent σ factor activating mechanism with Crl. In the Gram-positive bacterium, B. subtilis, YvrI-YvrHa are two small proteins that promote transcription of genes required in acidic conditions (87). Activation of gene transcription by the σ factor-like protein, YvrI, requires binding of its N terminus to the small co-regulator, YvrHa. In this system, YvrI is most similar to σ 4 domain, interacts with the β-flap domain of E, an interaction that is not dependent on YvrHa, and determines the promoter specificity of Eσ YvrI. YvrHa interacts with the β' subunit of RNAP, a conserved interaction for the σ 2 domain, and aids in open complex stabilization. As these independent proteins co-purify with RNAP from B. subtilis, YvrI and YvrHa likely function in vivo as a unified σ factor (46). ComW and σ X weakly interact in yeast-two-hybrid assays (37), but a direct interaction between these protein partners has not been demonstrated in competent pneumococcal cells or in vitro. Yet σ X is destabilized in Δ comW cells, or cells with N- or C-terminally tagged ComW (35), suggesting that the σ X-ComW interaction is important for stability of both proteins. ComWΔ6 levels are decreased compared with WT, resulting in decreased transformation. In addition, point mutations to ComW's electro-negative surface results in decreased transformation and decreased protein levels near peak competence. Thus it is possible that the C-terminal domain and/or the electro-negative surface of ComW mediate interaction with σ X. Interestingly, point mutations to ComW's DNA binding surface, specifically residues L42 and R44, disrupt transformation but produce stable protein. So it is likely that interaction with σ X is not mediated by this surface. Thus, it will be valuable to determine σ X protein levels in all ComW mutants in order to identify residues important for σ X stabilization. The σ X and ComW pair shares some similarity to B. subtilis' YvrI-YvrHa. At the protein level, σ X and YvrI are ∼23 kDa, and ComW and YvrHa are ∼10 kDa. SigX certainly functions as a σ factor (12), and ComW's predicted similarity to σ 2 domains suggests that it too can interact with E. Investigation into endogenous ComW protein-protein interactions are required to determine how it functions in the context of the Eσ X complex at competence onset. Orthologs of comW have been identified in anginosus and mitis group streptococci, and in the Gram-positive coccus, P. lacrimalis. All known ComWs are similar in primary structure and Robetta models of orthologs from S. anginosus and S. pseudopneumoniae are similar to that of pneumococcus. However, our findings that these orthologs differ in their ability to complement pneumococcal ComW during transformation and the observed differences in protein stability hint that species-specific determinants exist that dictate ComW function in vivo. As subunits of E are highly conserved among Gram-positive and Gram-negative organisms (47, 64), it is more likely that molecular interactions with species specific σ X are more important for ComW function. Like ComW, the σ X primary sequence is highly similar across species, but does vary. Furthermore, streptococci of different species groups utilize different quorum sensing systems to activate production of σ X and ComW seems to be unique to only one of these systems (ComCDE). If streptococci have evolved multiple mechanisms to regulate competence, the σ X-ComW system provides a unique opportunity for deep study into the co-evolution of transcriptional regulators. Lastly, although specific affinities of pneumococcal σ A and σ X for E have not yet been determined, σ A mutations that are predicted to disrupt Eσ A formation (mutations in σ 2 and σ 4 domains), bypass the ComW requirement (36, 37). Thus, competition for E between σ A and σ X at the onset of competence is likely a key determinant of transformation efficiency. It is conceivable ComW helps to tip the competition in favor of Eσ X formation over that of Eσ A. Previous work demonstrated that replacement of pneumococcal σ A with a chimeric σ A, which included S. mutans' regions σ 2 and σ 4, partially restored Δ comW phenotypes (unpublished, Tovpeko Thesis, 2016). Recall that S. mutans are naturally transformable streptococci that do not produce ComW during competence. This is additional evidence that some species-specific determinants exist and are important factors in streptococcal transformation. More broadly, species of the bovis, pyogenic, salivarius, and suis streptococcal groups likely evolved ComW-independent mechanisms to regulate σ X activity, and consequently, natural transformation. The pneumococcal competence specific ComW and σ X regulators provide a unique framework to study multiple molecular phenomena. ComW, predicted as structurally similar to σ-factors, and σ X, the only known σ-factor in streptococci, work together to promote a robust change in pneumococcal transcription during competence. They have been shown to directly interact with DNA and/or RNA polymerase, and both are required for efficient pneumococcal transformation. Taken together, these observations suggest that ComW and σ X function together, perhaps, for example, as a competence specific two-part σ factor. Therefore ComW and σ X provide an opportunity to study σ factor function, regulation, and holoenzyme assembly. ComW and σ X also offer an unambiguous context to study how independent proteins function together to promote gene expression. Lastly, further investigation into the ComW-σ X system will add insight into a mechanism used to control natural genetic transformation in an important human pathogen. This knowledge will add to the growing number of mechanisms that bacteria use to promote horizontal gene transfer. A deeper understanding of the mechanisms bacteria use to scavenge evolutionarily favorable genes will increase ability to combat their rapid adaptation to antibiotics and bacterial specific vaccines. Materials and methods Bacterial strains and culture media The bacterial strains used in this study are listed in Table S1. CP2137, a Δ cps Δ comA Δ ssbB::pEVP3::ssbB derivative of strain Rx1 (10, 36, 88) was used as the wildtype (WT) standard for transformation assays. CP2137 does not secrete endogenous competence stimulating peptide (CSP) due to deletion of the CSP exporter, ComA. All comW mutations were placed in the CP2137 background (see below). CP2463, a Δ comW::kan derivative of CP2137, was used as the Δ comW standard for transformation assays (36). All pneumococcal strains were cultured in CAT medium and plated on CAT or THY supplemented with 1.5% agar and selective antibiotic, as needed. CAT medium was prepared from 5 g of tryptone (Difco Laboratories), 10 g of enzymatic casein hydrolysate (Sigma), 1 g of yeast extract (Difco), and 5 g of NaCl (Fisher Scientific) in 1 liter of H 2 O, sterilized for 40 min at 121 °C, and then supplemented to 0.2% glucose and 0.016 m K 2 HPO 4 before use. THY was prepared from 10 g of yeast extract (Difco), 30 g of Todd Hewitt Broth (Difco) in 1 liter of H 2 O and sterilized for 20 min at 121 °C. Novobiocin was used in pneumococcal transformation assays at 2.5 μg/ml. E. coli strains DH5α and BL21De3 were hosts for plasmid isolation and protein expression, respectively. For plasmid introduction, E. Coli strains were chemically transformed according to (89). E. coli strains were cultured in lysogeny broth (LB) (90). LB was prepared from 5 g bacto tryptone (Difco), 5 g NaCl (Fisher Scientific), 2.5 g of yeast extract (Difco) in 1 liter H 2 O, sterilized for 20 min at 121 °C, and supplemented with appropriate antibiotics and 1.5% agar, as needed. Ampicillin was used at 100 μg/ml, for growth of E. coli strains. Antibiotics were purchased from Sigma. Computational modeling The ab initio structural models of pneumococcal ComW were calculated using the Robetta server (53), and with a local installation of Rosetta (49). The Robetta server generated 3-residue and 9-residue fragment files that were used as input for the Rosetta calculation. For the Rosetta calculation, radius of gyration, contact-order, and sheet filters were used. Helix and loop structures were weighted equally and a fast relax protocol was performed. Clustering of the 8,000 most energetically favorable models was done using Rosetta's cluster application with automatic radius detection. The Global Distance Test (Rosetta specific gdtmm) was used to cluster the models. Five clusters, each with nine models, were generated. One model from the five clusters was selected as representative of pneumococcal ComW's structure, based on Rosetta energy score, and cluster agreement as determined by rmsd between superimposed structures. Plasmid construction and procurement Plasmids and primers used in this study are listed in Tables S2 and S3, respectively. The comW truncation and orthologous variants in pET22b+ vectors were synthesized by Genscript (New Jersey). Gene sequences for comW were taken from the genomes of streptococcal species (NCBI): S. anginosus (ALJO01000004.1), S. oralis (NZ_NCUJ01000012.1), S. mitis (NZ_RMVN01000010.1), S. pneumoniae R6 (NC_003098.1) S. pseudopneumoniae (CP002925.1). Oligonucleotides were synthesized by IDT (Coralville, Iowa). Site directed mutagenesis of pNLI37 was used to generate point mutations in pneumococcal comW. pNLI37 was used as a template for PCR amplification by Platinum Pfx Polymerase (Thermo Fisher Scientific). Mutation specific primers (Table S3) and PCR conditions optimized for each primer set (individual conditions not given) in 50 μl reactions. PCR products were digested with DpnI (New England Biolabs) for 2 h at 37 °C. To circularize plasmids, a modified Seamless Ligation Cloning Extract (SLiCE) protocol (91) was used. Briefly, 8 μl of DpnI digested PCR product was mixed with 1.5 μl of SLiCE reaction buffer and 1.5 μl of SLiCE cell extract, and brought to 15 μl volume with H 2 O. Reaction mixtures were incubated at 37 °C for 15 min in a thermocycler. 3 μl SLiCE reaction products were transformed into 50 μl of DH5α cells (Invitrogen), and then plated on LB agar with 100 μg/ml ampicillin. Single colonies were picked and cultured in LB medium overnight at 37 °C. Plasmids were isolated using the ZymoPURE Plasmid Miniprep Kit (Zymo Research) and concentration was measured using a ND-1000 Spectrophotometer (Nanodrop). Mutations were confirmed by Sanger DNA sequencing (University of Illinois at Chicago Sequencing Core (UICSQC)). Gibson assembly (92) of pET22b+ containing comW point mutants was achieved by use of the Hifi Assembly Mix (New England Biolabs). Briefly, comW point mutants from plasmids pNLI80, pNLI87, pNLI88, pNLI89, pNLI90, pNLI91, and pNLI100 were PCR amplified using primer pair NL226 and NL233 which deleted residues 73 RGFISC 78 of comW. The PCR products contained 5′ and 3′, 12 bases overlap with NLI60 vector backbone that contained a C-terminal V5 epitope-6His tag (V5H6), followed by a stop codon. The vector backbone was PCR amplified using primer pair NL190 and NL232. PCR products were purified using the DNA Clean & Concentrator-5 kit (Zymo Research) and DNA concentrations measured using a Nanodrop spectrophotometer. DNA size and concentration was used to estimate picomolar concentration of each clean PCR product. Vectors and inserts were mixed at a 1:2 molar ratio, with up to 0.2 pmol of DNA in 10-μl reactions, and incubated for 15 min at 50 °C in a thermocylcer. Assembly products (3 μl) were transformed into DH5α cells, and plated on LB-agar plates with ampicillin overnight at 37 °C. Single colonies were picked and cultured in LB medium overnight at 37 °C. Plasmids were isolated and sequences confirmed with Sanger sequencing (UICSQC). The plasmids were transformed into BL21De3 (Invitrogen) cells and plated on LB-agar supplemented with 100 μg/ml ampicillin, overnight at 30 °C. Design and construction of DNA donors for pneumococcal comW gene replacement, and of the novobiocin resistance gene cassettes Strains, plasmids, and primers are listed in supporting information (Tables S1–S3). Strains CP2800, CP2803, and CP2805 were generated using restriction digestion and ligation. The flanking upstream (primers NL106 and NL107) and downstream (NL102 and NL103) regions of pneumococcal comW were PCR amplified with primers containing 5′ BtsI restriction sites using 5 ng of CP2137 genomic DNA, 1 μl of Phire Hot Start II DNA Polymerase (Thermo Fisher Scientific), and 0.2 m m dNTP mix (Thermo Fisher Scientific) in a 50-μl reaction in a thermocycler. The comW Δ 6 gene variant was PCR amplified using primer pair NL104 and NL105, Phire Hot Start II DNA Polymerase, 5 ng of pNLI60, and 0.2 m m dNTP mix in a 50-μl reaction, generating a fragment with 5′ and 3′ BtsI cut sites. The genomic upstream flank that was ligated to comW Δ 6 was generated using primer pair NL101 and NL106, 50 ng of CP2137 genomic DNA, and the same PCR conditions as primer pair NL106 and NL107. The E19A and Y41A mutations in comW were PCR amplified from plasmids pNLI88 and pNLI80, respectively, using primer pair NL105 and NL108, in the same PCR conditions as comW Δ 6 variant. To link cassette pieces, 10 μl of the flanking arms and comW variant PCR reactions were digested with BtsI (New England Biolabs) overnight at 37 °C in 30-μl reactions, to generate 2-bp overhangs on the ends of each molecule. Digested DNA products were purified using Zymo research kits and the DNA concentration was taken using a nanodrop. A 1:1 flank to gene fragment ratio was used in 30-μl ligation reactions overnight at 16 °C in a thermocycler. Ligation products were transformed into pneumococcus for homologous recombination (see below). Strains CP2801-CP2802, CP2804, CP2806-CP2807, and CP2808-CP2813 were obtained using Gibson assembly (92) with NEBuilder HiFi Assembly Mix (New England Biolabs). For assembly of pneumococcal comW point mutant cassettes, 5 ng of CP2137 genomic DNA was used in PCR with 0.1 units Phusion High-fidelity DNA polymerase (Thermo Fisher Scientific), 1 m m MgCl 2, 0.4 m m dNTP mix, and primers NL103 and NL177 to amplify a 3,327-bp comW downstream flank. Pneumococcal comW mutants were PCR amplified using 5 ng of specific plasmids with primer pair NL175 and NL176 containing 12-bp overlapping sequence with the upstream and downstream comW flanks in the PCR conditions mentioned above. 50 ng of CP2137 genomic DNA with primers NL179 and NL180 was used to amplify a 2,301-bp comW upstream flank, in the PCR conditions mentioned above without the use of MgCl 2. For assembly of comW orthologous gene cassettes, the flanking upstream (primers NL153 and NL154) and downstream (primers NL155 and NL156) regions of pneuococcal comW were PCR amplified using 50 ng of CP2137 genomic DNA, 0.02 units of Phusion High-fidelity DNA Polymerase, 0.4 m m dNTP mix, and 10 m m MgCl 2 in a 50-μl reaction in a thermocycler. Orthologous comW genes were amplified from plasmids (Table S2) with gene specific primers (Table S3) using 5 ng of plasmid DNA, 0.02 units of Phusion High-Fidelity DNA polymerase, 0.4 m m dNTP mix in 50-μl reactions, in a thermocycler. All PCR products were purified using Zymo kits. A Nanodrop was used to determine the concentration. Concentrations and DNA fragment sizes were used to calculate the number of picomole ends. Flanking arms and comW variant fragments were mixed in a 6:1 molar ratio and incubated in 10-μl reactions with HiFi assembly mix at 50 °C for 20 min in a thermocycler. 1 μl of each assembly product was used in subsequent PCR with primers DAM497 and DAM500 for amplification to generate gene cassette products of 2,200 bp. PCR products were purified and sequenced at the UICSQC. The 7.4-kb gyrB novobiocin resistance marker was prepared by amplification with primers YT76 and YT77, and 10 ng of either CM6 or CP1500 genomic DNA as the template. PCR amplification was performed in 50 μl reactions with 1 μl Phire HotStart II polymerase and Phire reaction buffer, and 0.2 m m of dNTP. The concentration of the purified product was measured in a ND-1000 spectrophotometer. Construction of comW mutant strains CP2137 cells were cultured in 10 ml of CAT medium to an OD 550 of 0.05. For transformation, 100 μl of cells were mixed with 100 ng of CSP, 0.04% BSA, 0.001 m CaCl 2, and 100 ng of gene replacement cassettes made by Gibson assembly, or 300 ng of gene replacement cassettes made by restriction digestion and ligation. Transformation reactions were brought to a total volume of 1 ml with CAT, and incubated with CSP and DNA for 2 h at 37 °C, and then chilled. Serial dilutions of each reaction were plated on CAT agar and grown overnight at 37 °C. Single colonies were picked and cultured overnight in CAT supplemented with 0.016 m K 2 HPO 4 at 37 °C. Overnight cultures were diluted into 10 ml CAT supplemented with 0.016 m K 2 HPO 4 and 0.2% glucose, and grown at 37 °C to an OD 550 of 0.2. Cells were diluted 1:5 in H 2 O and used in PCR reactions to amplify the comW region with primers DAM145 and DAM146. PCR products were cleaned and sequenced at the UICSQC. Pneumococcal transformation assays Aliquots of frozen pneumococcal strain stocks were diluted 1:100 in 10 ml of CAT medium and cultured at 37 °C for five hours until their OD 550 measured between 0.2–0.4. Each culture was diluted in CAT to an OD 500 0.05, and 10 μl of cells were incubated at 37 °C for 80 min with 100 ng CSP, 0.04% BSA, 0.001 m CaCl 2, and 100 ng novobiocin resistance cassette in 1 ml with CAT medium. Transformation reactions were then diluted 1:100 in THY medium and incubated at 37 °C for an additional 80 min, and then on ice prior to serial dilution. Diluted cells were plated on THY agar with 2.5 μg/ml novobiocin and incubated overnight at 37 °C. Colonies were counted and the transformation efficiency determined as a ratio of transformants/total cfu (CFU). ComWΔ6 purification E. coli transformed with a truncated variant of comW (ComWΔ6) in the pET22b+ vector were cultured in 3 L of LB medium at 37 °C, 200 rpm to an OD 600 0.5; then protein expression was induced by addition of [1 m m]f isopropyl β-d-1-thiogalactopyranoside (IPTG) (Gold Biotechnology). Induced cultures were incubated overnight at 20 °C, 200 rpm. The next day, the OD of each culture was taken using a spectrophotometer, and cells collected at 5,000 rpm, 4 °C, for 30 min. Cells were resuspended in 50 ml of E. coli resuspension buffer (50 m m Tris-HCl, pH 8.0, 500 m m NaCl) supplemented with 25 m mimidazole, 50 m m MgCl 2, 100 μg/ml DNase, and a protease inhibitor tablet (Roche Applied Science). The cell resuspension was lysed using an emulsiflex. The soluble fraction was separated by centrifugation at 30,000 × g for 30 min at 4 °C. The soluble supernatant was passed through a 5-ml nickel-nitrilotriacetic acid resin (Thermo Fisher Scientific) on a column equilibrated with Column Buffer (50 m m Tris-HCl, pH 8.0, 500 m m NaCl, 10% glycerol) via gravity. The column was washed with 500 ml of Column Buffer supplemented with 5 m m β-mercaptoethanol (βME) and 40 m m imidazole. ComWΔ6 was eluted from the column with 10 ml of column buffer supplemented with 5 m m BME and 275 m m imidazole. The ComWΔ6 eluate was dialyzed into 50 m m Tris-HCl, pH 8.0, 500 m m NaCl, 1 m m EDTA, 10% glycerol, and 1 m m βME. Protein concentration was measured using a Nanodrop. Nano differential scanning fluorimetry After purification, ComWΔ6 was diluted to 0.5 mg/ml in ten different biological buffers (Fig. S2). Diluted ComWΔ6 was loaded into 10-μl capillaries in duplicate. Each capillary was loaded into a Prometheus NT.48 instrument (NanoTemper Technologies Inc). Samples were heated from 20 °C to 95 °C with a temperature ramp of 1.0 °C/min. Data were analyzed with the PR.ThermControl software (NanoTemper Technologies Inc). Analytical ultra-centrifugation Purified ComWΔ6 was diluted to an absorbance (280 nm) of 1.5 (0.92 mg/ml), 1.0 (0.61 mg/ml), and 0.5 (0.31 mg/ml) in cold 50 m m Tris-HCl, pH 8.0, 500 m m NaCl, 1 m m EDTA, and 1 m m βME. Adjacent channels of analytical ultracentrifugation cells were separately loaded with 400 μl of ComWΔ6 sample or 420 μl of buffer and sealed under pressure. The samples were centrifuged at 40,000 rpm in a Beckman Coulter XL-I at 10 °C for 24 h. Scans were taken every minute. The raw sedimentation data were analyzed using a continuous distribution model with SEDFIT and then analyzed with a Bayesian model. Size-exclusion chromatography ComWΔ6 variants were concentrated in 50 m m Tris-HCl, pH 8.0, 500 m m NaCl, 1 m m EDTA, 10% glycerol, and 1 m m βME and applied to a pre-packed Superdex 200 column (GE Healthcare) on an AKTA (Amersham Biosciences). The proteins were eluted at 0.5 ml/min, and collected in 1 ml fractions for analysis on SDS-polyacrylamide gels. Small standards RNase (13.7 kDa), chymotrypsinogen A (25.6 kDa), Ovalbumin (43.0 kDa), and Conalbumin (75.0 kDa) were applied over the same column in the same conditions. Circular dichroism (CD) Purified ComWΔ6 variants were diluted to 0.02 mg/ml in 30 m m sodium fluoride (NaF) in a 0.2 cm path quartz cell. Their structure was analyzed at 25 °C from 260 nm to 190 nm using a JASCO J-710 machine. Deconvolution of the data were performed using Dichroweb using the Contin-LL method (93, 9495). Electrophoretic mobility shift assay EMSA probes were generated using PCR amplification. For amplification of S. pneumoniae's combox promoter region, upstream of ssbB, 50 ng of CP2137 genomic DNA was used as a template for amplification with primers NL223 (6′FAM) and NL224 for labeled probe, and NL228 and NL224 for unlabeled probe. For amplification of S. mutans ComR-box, 1 μl of S. mutans cells diluted 1:5 in H 2 O were used as a template for amplification with primers NE 25 (6′FAM) and NE27 for labeled probe, and NE26 and NE27 for unlabeled probe. Probes were PCR purified using Zymo kits and concentration measured using a nanodrop. Purified protein at varied concentrations, and 20 n m of DNA probe were mixed together in 5× EMSA reaction buffer (250 m m Tris-HCl, pH 8.0, 1.5 m NaCl, 5 m m MgCl 2, 5 m m EDTA), supplemented with 50 m m βME, and the volume of each reaction was brought to 20 μl with ComWΔ6 dialysis buffer plus 30% glycerol. After incubation at 25 °C for 30 min, in a thermocycler, wells of a 5% polyacrylamide, nondenaturing gel were loaded with 10 μl of each EMSA reaction and run at 90 V for 4 h at 4 °C in 1X TBE. Gels were removed from casting plates and imaged on Typhoon 3000 (GE Healthcare). Band intensities were measured using Image Studio Lite software (Li-COR biosciences), and the amount of probe shifted by protein was determined as (bound probe/total probe) for each lane. The gels shown are representative of three biological replicates, and statistical significance was determined using an unpaired student's t test. Western blotting Pneumococcal cells were cultured in 10 ml of CAT medium for 4 h, diluted to an OD 550 = 0.2, and induced to competence by addition of 100 ng of CSP, 0.04% BSA, and 0.001 m CaCl 2, for 17 min, and then chilled. A 10-μl sample of each culture was removed for plating on CAT agar and incubated overnight at 37 °C to determine CFU. The remaining cells were collect by centrifugation at 4,000 rpm for 30 min at 4 °C. After removal of the supernatant, cell pellets were resuspended in pneumococcal wash saline (0.01 m Tris-HCl, pH 8.0, 0.15 m NaCl, and 0.01 m EDTA), and 0.1 mm beads were added to each tube for lysis by bead beating in the cold for one 5-min interval, followed by one 4-min interval. The lysates were centrifuged for 10 s and 180 μl of clarified cell lysates were moved to a fresh 1.5-ml microcentrifuge tube with 60 μl of 4X Laemmli buffer (Bio Rad). Cell samples were incubated at 95 °C for 3 min and stored at −20 °C. For protein detection, the wells of a precast 4–20% SDS-PAGE (Bio Rad) gel (8.6 × 6.7 cm, 0.1 cm thickness) were filled with 1 × 10 8 cells of each sample. Gels were run at 200 V for 35 min. Proteins were then transferred for 2 h at 36 V to a PVDF membrane activated with cold methanol at 30 V at 4 °C in 1X transfer buffer (25 m m Tris-HCl, pH 8.0, 192 m m glycine, 10% MeOH). Membranes were then quickly rinsed with 1× TBST (20 m m Tris-HCl, pH 7.6, 137 m m NaCl, 0.1% Tween 20). Membranes were blocked with 5% BSA in 0.1% TBST for 1 h at room temperature with rocking. Blots were then incubated with a primary rabbit antibody, raised against pneumococcal ComW (48) (1:1000 in 1.0% BSA in 0.1% TBST), plus a primary mouse antibody raised against E. coli RNA polymerase β-subunit (Thermo Fisher Scientific) (1:3000 in 0.5% BSA in 0.1% TBST) overnight at 4 °C, with rocking. After three washes with 0.1% TBST for 5 min at room temperature, blots were simultaneously incubated with goat anti-rabbit and goat anti-mouse secondary antibodies (Sigma) diluted 1:10,000 in 1% BSA in 0.1% TBST, for 1 h and 20 min at room temperature, with rocking. After one 10-min wash and two 5-min washes in 0.1% TBST at room temperature, blots were incubated with 10 ml of ECL substrate (Bio Rad) for 5 min at room temperature, with rocking. The wet blots were imaged using the LiCOR system. Image Studio Lite was used to quantify the signal intensity from endogenous ComW bands (9.5 kDa) and RNA polymerase (137 kDa). Signal intensities were normalized to the RNA polymerase signal. The blots shown are representative of three biological replicates. Statistical significance was determined using an unpaired Student's t test. Author contributions N. L. I., G. P., and D. A. M. conceptualization; N. L. I., G. P., and D. A. M. resources; N. L. I. and G. P. data curation; N. L. I., G. P., and D. A. M. formal analysis; N. L. I., G. P., and D. A. M. funding acquisition; N. L. I., G. P., and D. A. M. investigation; N. L. I., G. P., and D. A. M. methodology; N. L. I. and G. P. writing-original draft; N. L. I., G. P., and D. A. M. writing-review and editing; G. P. and D. A. M. supervision; G. P. and D. A. M. project administration. Acknowledgments We thank The Center for Structural Biology at the University of Illinois at Chicago for use of the lab space equipment. 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The bacterial RNA polymerase core enzyme is a complex of five subunits that interacts transitory with one of a set of sigma factors forming the RNA polymerase holoenzyme. The sigma factor confers promoter specificity to the RNA polymerase. In the Gram-positive pathogenic bacterium Streptococcus pneumoniae, most promoters are likely recognized by SigA, a poorly studied housekeeping sigma factor. Here we present a sequence conservation analysis and show that SigA has similar protein architecture to Escherichia coli and Bacillus subtilis homologs, namely the poorly conserved N-terminal 100 residues and well-conserved rest of the protein (domains 2, 3, and 4). Further, we have purified the native (untagged) SigA protein encoded by the pneumococcal R6 strain and reconstituted an RNA polymerase holoenzyme composed of the E. coli core enzyme and the sigma factor SigA (RNAP-SigA). By in vitro transcription, we have found that RNAP-SigA was able to recognize particular promoters, not only from the pneumococcal chromosome but also from the S. agalactiae promiscuous antibiotic-resistance plasmid pMV158. Specifically, SigA was able to direct the RNA polymerase to transcribe genes involved in replication and conjugative mobilization of plasmid pMV158. Our results point to the versatility of SigA in promoter recognition and its contribution to the promiscuity of plasmid pMV158. ### Regulation of pneumococcal epigenetic and colony phases by multiple two-component regulatory systems 2020, Plos Pathogens Show abstract Streptococcus pneumoniae is well known for phase variation between opaque (O) and transparent (T) colonies within clonal populations. While the O variant is specialized in invasive infection (with a thicker capsule and higher resistance to host clearance), the T counterpart possesses a relatively thinner capsule and thereby higher airway adherence and colonization. Our previous study found that phase variation is caused by reversible switches of the “opaque ON-or-OFF” methylomes or methylation patterns of pneumococcal genome, which is dominantly driven by the PsrA-catalyzed inversions of the DNA methyltransferase hsdS genes. This study revealed that switch frequency between the O and T variants is regulated by five transcriptional response regulators (rr) of the two-component systems (TCSs). The mutants of rr06, rr08, rr09, rr11 and rr14 produced significantly fewer O and more T colonies. Further mutagenesis revealed that RR06, RR08, RR09 and RR11 enrich the O variant by modulating the directions of the PsrA-catalyzed inversion reactions. In contrast, the impact of RR14 (RitR) on phase variation is independent of PsrA. Consistently, SMRT sequencing uncovered significantly diminished “opaque ON” methylome in the mutants of rr06, rr08, rr09 and rr11 but not that of rr14. Lastly, the phosphorylated form of RR11 was shown to activate the transcription of comW and two sugar utilization systems that are necessary for maintenance of the “opaque ON” genotype and phenotype. This work has thus uncovered multiple novel mechanisms that balance pneumococcal epigenetic status and physiology. ### Type II bacterial toxin-antitoxins: hypotheses, facts, and the newfound plethora of the PezAT system 2023, FEMS Microbiology Reviews ### The Facts and Family Secrets of Plasmids That Replicate via the Rolling-Circle Mechanism 2022, Microbiology and Molecular Biology Reviews ### The Role of Streptococcus pneumoniae in Community-Acquired Pneumonia 2020, Seminars in Respiratory and Critical Care Medicine ### Environmental impact on differential composition of gut microbiota in indoor chickens in commercial production and outdoor, backyard chickens 2020, Microorganisms View all citing articles on Scopus Funding was provided by the University of Illinois at Chicago through the Abraham Lincoln Fellowship awarded to NLI, as well as an R03 grant to GP and DAM from the National Institute of Allergy and Infectious Diseases of the National Institutes of Health (1R03AI128228). The authors declare that they have no conflicts of interest with the contents of this article. This article contains Figs. S1–S6 and Tables S1–S3. 3 abbreviations used are: σ σ factor σ X SigX E core RNA polymerase enzyme Eσ core RNA polymerase-σ factor holoenzyme CSP competence stimulating peptide EMSA electrophoretic mobility shift assay βME β-mercaptoethanol. © 2019 Inniss et al. 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https://www.quora.com/How-can-we-prove-that-C-n-k-C-n-1-k-+C-n-1-k-1
How can we prove that C(n,k) = C (n-1,k) +C (n-1,k-1)? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Binomial Coefficients Combinatorics C Proofs (mathematics) Permutations and Combinat... Prove Mathematically Combinatorial Algebra Binomial Expansion Mathematical Proof 5 How can we prove that C(n,k) = C (n-1,k) +C (n-1,k-1)? All related (38) Sort Recommended Thomas Cormen The C in CLRS. · Upvoted by Yair Livne , Master's Mathematics, Hebrew University of Jerusalem (2007) · Author has 843 answers and 41M answer views ·10y Originally Answered: How can you prove (Pascal's formula) that C(n,k) = C (n-1,k) +C (n-1,k-1)? · You don't have to order the players in any particular way. You can just say "Choose one player. Let's call him or her Pat. You either choose Pat or you don't. The number of ways to choose k players out of n players is the sum of the ways that include Pat among the chosen players and the ways that don't include Pat. If you choose Pat, then you have to choose k-1 players from the remaining n-1 players. If you don't choose Pat, then you still have to choose k players from the remaining n-1 players." [I meant to enter this as a comment to Shane Barratt's answer. Oops.] Upvote · 999 286 9 7 Related questions More answers below How can you show that c(n,k)=c(n-1, k-2)+c(n-1, k-1)? How do I prove C (n, r) + C (n, r-1) = C (n+1, r)? How can one prove that (n+1)!/[(k-1)!(n-k)!] = (n+1)!/[k!(n-k+1)!]? How can I prove that C (n, 0)C (m, k) + C (n, 1)C (m, k - 1) + … + C (n, k-1)C (m, 1) + C (n, k)C (m, 0) = C (n + m, k) with binomial theorem? Note: C(x, y) = “x choose y” How do I prove this combinatorial identity C k+1 k+n+1=C k k+C k k+1+C k k+2+...+C k k+n C k+n+1 k+1=C k k+C k+1 k+C k+2 k+...+C k+n k? Assistant Bot · 1y The identity C(n,k)=C(n−1,k)+C(n−1,k−1)C(n,k)=C(n−1,k)+C(n−1,k−1) is known as Pascal's identity and can be proved using a combinatorial argument. Here's a step-by-step explanation of the proof: Combinatorial Proof: Understanding C(n,k)C(n,k): C(n,k)C(n,k) represents the number of ways to choose k k elements from a set of n n elements. Consider a Set of n n Elements: Let’s denote the n n elements in the set as {1,2,…,n}{1,2,…,n}. Choose k k Elements: When selecting k k elements from this set, there are two scenarios to consider based on whether the element n n is included in the selection or not. Case 1: Element n n is Included Continue Reading The identity C(n,k)=C(n−1,k)+C(n−1,k−1)C(n,k)=C(n−1,k)+C(n−1,k−1) is known as Pascal's identity and can be proved using a combinatorial argument. Here's a step-by-step explanation of the proof: Combinatorial Proof: Understanding C(n,k)C(n,k): C(n,k)C(n,k) represents the number of ways to choose k k elements from a set of n n elements. Consider a Set of n n Elements: Let’s denote the n n elements in the set as {1,2,…,n}{1,2,…,n}. Choose k k Elements: When selecting k k elements from this set, there are two scenarios to consider based on whether the element n n is included in the selection or not. Case 1: Element n n is Included: If n n is included in the selection, we need to choose the remaining k−1 k−1 elements from the first n−1 n−1 elements (i.e., {1,2,…,n−1}{1,2,…,n−1}). The number of ways to do this is C(n−1,k−1)C(n−1,k−1). Case 2: Element n n is Not Included: If n n is not included in the selection, we need to choose all k k elements from the first n−1 n−1 elements. The number of ways to do this is C(n−1,k)C(n−1,k). Combining the Cases: By the principle of counting, the total number of ways to choose k k elements from n n elements is the sum of the two cases: C(n,k)=C(n−1,k)+C(n−1,k−1)C(n,k)=C(n−1,k)+C(n−1,k−1) Conclusion: This combinatorial argument shows that the identity C(n,k)=C(n−1,k)+C(n−1,k−1)C(n,k)=C(n−1,k)+C(n−1,k−1) holds true, as it effectively counts the same selection process in two different ways. This identity is fundamental in combinatorics and forms the basis for constructing Pascal's Triangle, among other applications. Upvote · Shane Barratt EECS at Berkeley · Upvoted by Jay Wacker , theoretical physicist and David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 54 answers and 116.1K answer views ·10y Originally Answered: How can you prove (Pascal's formula) that C(n,k) = C (n-1,k) +C (n-1,k-1)? · Easiest way would be a combinatorial proof, where you prove that counting a certain set of items can be done by either the left or right hand side of the inequality. Left hand side: Choosing k basketball players out of n. Right hand side: Order the n basketball players by height. Each set will either have the tallest basketball player or it will not. C(n-1,k) means that set doesn't include the tallest basketball player, because you still have to choose k players. C(n-1,k-1) means you chose the tallest basketball player and only have k-1 to go. Therefore, LHS=RHS Upvote · 99 20 Raviphol Sukhajoti Software Engineer ·10y Originally Answered: How can you prove (Pascal's formula) that C(n,k) = C (n-1,k) +C (n-1,k-1)? · Another way to think of this is to use a lattice path of k rows, n-k columns. There are C(n, k) different path from bottom left to top right corner. If you restrict the first move to go up, then there are C(n-1, k-1) different paths. You will get C(n-1, k) for the case where you move right in the first step. Then you got C(n, k) = C(n-1, k) + C(n-1, k-1) = number of lattice paths of size k x (n-k). Upvote · 9 5 9 1 Related questions More answers below How do I show that C (n ,r) +C (n,n-r) =C (N+1,r)? How can I prove ∑n k=r C k r=C n+1 r+1∑k=r n C r k=C r+1 n+1 ? How can one prove that {n! / [k! (n-k)! ]} + {n! / [(k-1)! (n-k+1)! ]} = (n+1)! / [k! (n-k+1)! ]? How do I prove that n • C(n,r) = (r+1) • C (n,r+1) + r • C(n,r)? How do you show that c (n, r) =c (n - 1, r - 1) + c (n-1, r)? Umang Malhotra B.E. from Birla Institute of Technology, Mesra ·8y There are many ways to prove it. You may be looking for explanation for a specific way - in which case you need to post that way. Here is perhaps a simple proof - the LHS i.e. C(n,k) is the number of ways to select k objects from n distinct ones. Now let us say we have marked a specific object and want to count the selections with and without this object. You can select k objects without the marked one in C(n−1,k) ways, and including the marked one in C(n−1,k−1) ways - thus both these sum up to the LHS Upvote · 9 3 9 2 Anirban Ghoshal Programmer by profession. · Author has 2.1K answers and 5.8M answer views ·10y This is simple to prove directly: C(n−1,k)+C(n−1,k−1)C(n−1,k)+C(n−1,k−1) =(n−1)!k!(n−k−1)!+(n−1)!(k−1)!(n−k)!=(n−1)!k!(n−k−1)!+(n−1)!(k−1)!(n−k)! =(n−k)(n−1)!+k(n−1)!k!(n−k)!=(n−k)(n−1)!+k(n−1)!k!(n−k)! =n!(n−k)!k!=C(n,k)=n!(n−k)!k!=C(n,k) Upvote · 9 7 9 2 Nigus Bantaymol 6y bowel contain 10 red and 10 blue ball a boy select balls at random with out look at them ,how many balls he selected to be sure of having at least three balls of the same color? Upvote · Him Author has 1.5K answers and 478.7K answer views ·1y Originally Answered: What is the combinatorial proof for C(n, k) = C(n - 1, k-1)+C(n-1,k)? · The LHS is clearly the number of ways to choose k things out of n things. Now, alternatively, we can fix one object out of n objects. Then, it only remains to select (k-1) objects out of (n-1) objects. This can be done in C(n-1,k-1) ways. Now it remains to count the combinations where the aforementioned fixed object isn't there. Which can be done in C(n-1,k) ways. Adding those, C(n,k) = C(n-1,k-1) + C(n-1,k) Upvote · Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Upvoted by Richard Shearer , MSc (Biochemistry) Biochemistry & Mathematics, University of the Witwatersrand · Author has 6.8K answers and 52.8M answer views ·4y Related How do I prove C (n, r) + C (n, r-1) = C (n+1, r)? These are the three basic expressions. All we need to do now is factorise and simplify the numerators: Quod erat demonstrandum Continue Reading These are the three basic expressions. All we need to do now is factorise and simplify the numerators: Quod erat demonstrandum Upvote · 99 14 9 8 Daniel Claydon Learning mathematics · Upvoted by Jeremy Collins , M.A. Mathematics, Trinity College, Cambridge and Aditya Garg , M.Sc. Mathematics, Indian Institute of Technology, Delhi (2013) · Author has 779 answers and 4.3M answer views ·6y Related How can I prove ∑n k=r C k r=C n+1 r+1∑k=r n C r k=C r+1 n+1 ? I’ll give you four proofs. Proof 1: Combinatorial Argument The right hand side is (n+1 r+1)(n+1 r+1), which represents the number of ways to pick r+1 r+1 elements from a set of n+1 n+1. Counting another way, imagine you had n+1 n+1 items lined up and you want to pick r+1 r+1. Let’s say the final item you pick is in position k k (for r+1≤k≤n+1 r+1≤k≤n+1). Then there are k−1 k−1 prior positions from which you pick the remaining r r items in (k−1 r)(k−1 r) ways. Summing across all k k shows (n+1 r+1)=n+1∑k=r+1(k−1 r)(n+1 r+1)=∑k=r+1 n+1(k−1 r) and shifting the index down by one gives the required result. Pro Continue Reading I’ll give you four proofs. Proof 1: Combinatorial Argument The right hand side is (n+1 r+1)(n+1 r+1), which represents the number of ways to pick r+1 r+1 elements from a set of n+1 n+1. Counting another way, imagine you had n+1 n+1 items lined up and you want to pick r+1 r+1. Let’s say the final item you pick is in position k k (for r+1≤k≤n+1 r+1≤k≤n+1). Then there are k−1 k−1 prior positions from which you pick the remaining r r items in (k−1 r)(k−1 r) ways. Summing across all k k shows (n+1 r+1)=n+1∑k=r+1(k−1 r)(n+1 r+1)=∑k=r+1 n+1(k−1 r) and shifting the index down by one gives the required result. Proof 2: Telescoping Series It is well known that (n r)=(n−1 r)+(n−1 r−1)(n r)=(n−1 r)+(n−1 r−1), so we have (k r)=(k+1 r+1)−(k r+1)(k r)=(k+1 r+1)−(k r+1). Then the left hand side may be written as n∑k=r(k+1 r+1)−(k r+1)=(n+1 r+1)∑k=r n(k+1 r+1)−(k r+1)=(n+1 r+1) as required. Note that we used the result (r r+1)=0(r r+1)=0. Proof 3: Induction The base case can be proven readily enough — when n=r n=r both sides are just 1 1. Then we assume ∑n k=r(k r)=(n+1 r+1)∑k=r n(k r)=(n+1 r+1). Under this assumption, n+1∑k=r(k r)=(n+1 r+1)+(n+1 r)=(n+2 r+1)∑k=r n+1(k r)=(n+1 r+1)+(n+1 r)=(n+2 r+1) using the same identity as in proof 2. This completes the inductive step and the result follows. Proof 4: Equating Coefficients Consider the polynomial p(x)=1+(1+x)+(1+x)2+⋯+(1+x)n p(x)=1+(1+x)+(1+x)2+⋯+(1+x)n Then the coefficient of x r x r in each expression (1+x)k(1+x)k is (k r)(k r) so the coefficient of x r x r in p(x)p(x) is ∑n k=r(k r)∑k=r n(k r). Alternatively, one may sum p(x)p(x) as a geometric series to get p(x)=(1+x)n+1−1 x p(x)=(1+x)n+1−1 x the coefficient of x r x r in this is just the coefficient of x r+1 x r+1 in (1+x)n+1(1+x)n+1, which is (n+1 r+1)(n+1 r+1), and the result follows. Upvote · 99 45 9 2 9 3 Amitabha Tripathi have been teaching Discrete Mathematics for almost 40 years · Author has 4.7K answers and 13.9M answer views ·4y Related How do I prove this combinatorial identity C k+1 k+n+1=C k k+C k k+1+C k k+2+...+C k k+n C k+n+1 k+1=C k k+C k+1 k+C k+2 k+...+C k+n k? We give three standard proofs of the identity (k k)+(k+1 k)+(k+2 k)+⋯+(k+n k)=(k+n+1 k+1)(k k)+(k+1 k)+(k+2 k)+⋯+(k+n k)=(k+n+1 k+1). …(⋆)…(⋆) First proof. ((combinatorial argument)) The RHS in (⋆)(⋆) represents the number of selections of k+1 k+1 elements from {1,…,k+n+1}{1,…,k+n+1}. The largest element in each selection must lie in {k+1,…,k+n+1}{k+1,…,k+n+1}. If k+i k+i denotes the largest element selected, so that 1≤i≤n+1 1≤i≤n+1, the remaining k k selections must come from {1,…,k+i−1}{1,…,k+i−1}. These selections can be done in (k+i−1 k)(k+i−1 k) ways, so that the total number of selectio Continue Reading We give three standard proofs of the identity (k k)+(k+1 k)+(k+2 k)+⋯+(k+n k)=(k+n+1 k+1)(k k)+(k+1 k)+(k+2 k)+⋯+(k+n k)=(k+n+1 k+1). …(⋆)…(⋆) First proof. ((combinatorial argument)) The RHS in (⋆)(⋆) represents the number of selections of k+1 k+1 elements from {1,…,k+n+1}{1,…,k+n+1}. The largest element in each selection must lie in {k+1,…,k+n+1}{k+1,…,k+n+1}. If k+i k+i denotes the largest element selected, so that 1≤i≤n+1 1≤i≤n+1, the remaining k k selections must come from {1,…,k+i−1}{1,…,k+i−1}. These selections can be done in (k+i−1 k)(k+i−1 k) ways, so that the total number of selections is ∑n+1 i=1(k+i−1 k)=∑n i=0(k+i k)∑i=1 n+1(k+i−1 k)=∑i=0 n(k+i k), which is the LHS in (⋆)(⋆). ■◼ Second proof.((using binomial identity)) Summing the binomial identity (n+1 r)=(n r)+(n r−1)(n+1 r)=(n r)+(n r−1) gives n∑i=0((k+i+1 k+1)−(k+i k+1))=n∑i=0(k+i k)∑i=0 n((k+i+1 k+1)−(k+i k+1))=∑i=0 n(k+i k). …(1)…(1) The LHS in (1)(1) telescopes to (k+n+1 k+1)−(k k+1)=(k+n+1 k+1)(k+n+1 k+1)−(k k+1)=(k+n+1 k+1). The RHS in (1)(1) is the RHS in (⋆)(⋆). ■◼ Third proof. ((using the binomial theorem)) We have 1+(1+x)+(1+x)2+⋯+(1+x)k+n=(1+x)k+n+1−1(1+x)−1 1+(1+x)+(1+x)2+⋯+(1+x)k+n=(1+x)k+n+1−1(1+x)−1, so that x(1+(1+x)+(1+x)2+⋯+(1+x)k+n)=(1+x)k+n+1−1 x(1+(1+x)+(1+x)2+⋯+(1+x)k+n)=(1+x)k+n+1−1. …(2)…(2) The coefficient of x k+1 x k+1 in the RHS of (2)(2) is the RHS in (⋆)(⋆). The coefficient of x k+1 x k+1 in the LHS of (2)(2), which is also the coefficient of x k x k in 1+(1+x)+(1+x)2+⋯+(1+x)k+n 1+(1+x)+(1+x)2+⋯+(1+x)k+n, is the LHS in (⋆)(⋆). ■◼ Upvote · 99 13 Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·6y Related How do you prove C (n + 1, r + 1) = C (n, r) + C (n, r + 1)? The key idea in this proof is realising that, for example, 7 × 6! = 7! or algebraically, (b + 1) × (b!) = (b + 1)! I will ... Upvote · 9 6 Joe Blitzstein Stanford PhD in Mathematics · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) and Jay Wacker , theoretical physicist · Author has 262 answers and 4M answer views ·10y Related What is an intuitive explanation for (n k)=(n−1 k−1)+(n−1 k)(n k)=(n−1 k−1)+(n−1 k)? See the story proofs section of Stat 110 Strategic Practice 1 at Page on harvard.edu. A nice aspect of story proofs is that they not only are proofs, but also they explain the intuition and logic of what's going on. In particular, the idea for seeing via story why (n k)+(n k−1)=(n+1 k)(n k)+(n k−1)=(n+1 k) is to imagine that you are in a club with n+1 people (including you). A committee of size k needs to be chosen. Then consider the two possible types of committee, based on whether or not you are on the committee. Upvote · 999 232 99 12 Related questions How can you show that c(n,k)=c(n-1, k-2)+c(n-1, k-1)? How do I prove C (n, r) + C (n, r-1) = C (n+1, r)? How can one prove that (n+1)!/[(k-1)!(n-k)!] = (n+1)!/[k!(n-k+1)!]? How can I prove that C (n, 0)C (m, k) + C (n, 1)C (m, k - 1) + … + C (n, k-1)C (m, 1) + C (n, k)C (m, 0) = C (n + m, k) with binomial theorem? Note: C(x, y) = “x choose y” How do I prove this combinatorial identity C k+1 k+n+1=C k k+C k k+1+C k k+2+...+C k k+n C k+n+1 k+1=C k k+C k+1 k+C k+2 k+...+C k+n k? How do I show that C (n ,r) +C (n,n-r) =C (N+1,r)? How can I prove ∑n k=r C k r=C n+1 r+1∑k=r n C r k=C r+1 n+1 ? How can one prove that {n! / [k! (n-k)! ]} + {n! / [(k-1)! (n-k+1)! ]} = (n+1)! / [k! (n-k+1)! ]? How do I prove that n • C(n,r) = (r+1) • C (n,r+1) + r • C(n,r)? How do you show that c (n, r) =c (n - 1, r - 1) + c (n-1, r)? How do I prove that n • C(n,r) = r • C(n,r) + C (n,r+1)? Why is C(n,k) =C (n,n-k)? How can you simplify (C stands for choose) 1 (n C k) +2 ((n-1) C k) +3 (n-2 C k) +4 (n-3 C k) +…+ (n-1-k) (k C k)? How can you prove that, C(n,r) = C (n-1,r-1) +C (n-2,r-1) +C (n-3,r-1) + …+ C (r-1,r-1) given that r > 1? How do you prove that 3 divides (3^n) C(k) for 1≤k<3^n? Related questions How can you show that c(n,k)=c(n-1, k-2)+c(n-1, k-1)? How do I prove C (n, r) + C (n, r-1) = C (n+1, r)? How can one prove that (n+1)!/[(k-1)!(n-k)!] = (n+1)!/[k!(n-k+1)!]? How can I prove that C (n, 0)C (m, k) + C (n, 1)C (m, k - 1) + … + C (n, k-1)C (m, 1) + C (n, k)C (m, 0) = C (n + m, k) with binomial theorem? Note: C(x, y) = “x choose y” How do I prove this combinatorial identity C k+1 k+n+1=C k k+C k k+1+C k k+2+...+C k k+n C k+n+1 k+1=C k k+C k+1 k+C k+2 k+...+C k+n k? How do I show that C (n ,r) +C (n,n-r) =C (N+1,r)? How can I prove ∑n k=r C k r=C n+1 r+1∑k=r n C r k=C r+1 n+1 ? How can one prove that {n! / [k! (n-k)! ]} + {n! / [(k-1)! (n-k+1)! ]} = (n+1)! / [k! (n-k+1)! ]? How do I prove that n • C(n,r) = (r+1) • C (n,r+1) + r • C(n,r)? How do you show that c (n, r) =c (n - 1, r - 1) + c (n-1, r)? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025 Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. 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In the following question, out of the given four alternatives, select the one which is opposite in meaning of the given word.Dank Hello, Guest Login / Register All Categories+ Test SeriesQuizzesPrevious Year PapersLive TestsLive QuizzesCurrent AffairsVideosNewsContact Us Get App All Exams Test series for 1 year @ ₹349 onlyEnroll Now ×× Home English Synonyms or Antonyms in the following question out of the given four al Question English In the following question, out of the given four alternatives, select the one which is opposite in meaning of the given word. Dank Dry Close Sticky Moist Solution The correct answer is Dry Finding the Opposite of 'Dank' The question asks us to find the word that is opposite in meaning to the given word, 'Dank'. To do this, we first need to understand the meaning of 'Dank' and then look at the meanings of the provided options. Understanding the Word 'Dank' The word 'Dank' is typically used to describe something that is unpleasantly damp, musty, and cold. It often refers to places like cellars, caves, or rooms that lack ventilation and sunlight, leading to a feeling of dampness and sometimes a stale smell. Dank: Unpleasantly damp and cold. Analyzing the Options Let's examine the meaning of each option provided: | Option | Meaning | --- | | Dry | Free from moisture or liquid; not wet or moist. | | Close | Near in space or time; only a short distance away; shut tightly. | | Sticky | Tending or appearing to stick to a surface or substance; gluey. | | Moist | Slightly wet; damp. | Identifying the Antonym (Opposite Meaning) We are looking for the opposite of 'Dank', which means unpleasantly damp and cold. Let's compare 'Dank' to the options: 'Dry' means free from moisture. This is the direct opposite of being damp or moist. 'Close' relates to distance or being shut, which is not related to dampness. 'Sticky' relates to having adhesive properties, which is also unrelated to dampness. 'Moist' means slightly wet or damp. This is similar in meaning to 'Dank' (although 'Dank' often implies an unpleasant quality), not opposite. Comparing the meanings, 'Dry' is the clear opposite of 'Dank'. Conclusion Based on the analysis of the meanings, the word that is opposite in meaning to 'Dank' is 'Dry'. Revision Table: Key Vocabulary | Word | Meaning | Antonym (Opposite) | Synonym (Similar) | --- --- | | Dank | Unpleasantly damp and cold | Dry | Moist, damp (often with negative connotation) | | Dry | Free from moisture | Dank, Wet, Moist | Arid, Parched | | Moist | Slightly wet; damp | Dry | Damp, Humid | Additional Information: Antonyms and Synonyms Understanding antonyms and synonyms is crucial for building a strong vocabulary. Antonyms are words that have opposite meanings, while synonyms are words that have similar meanings. Learning these relationships between words helps improve reading comprehension and writing skills. Antonyms: Words like 'hot' and 'cold', 'up' and 'down', 'light' and 'dark' are antonyms. Synonyms: Words like 'happy' and 'joyful', 'big' and 'large', 'quick' and 'fast' are synonyms. Context is also important when choosing synonyms or antonyms, as some words might be similar or opposite in one context but not another. Download PDF Was this answer helpful? 0 0 Important Questions from Synonyms or Antonyms Select the most appropriate ANTONYM of the given word. Contentment English View Answer 2. Select the most appropriate synonym of the given word. Apprehension English View Answer 3. Identify the ANTONYM of the following word in the given sentence. Divulge The culprit surprised everyone by opting to conceal the secret execution of his plans even after getting beaten by the police. English View Answer 4. Select the most appropriate ANTONYM of the given word. Adipose English View Answer 5. Which of the following options is the closest in meaning to the word underlined in the sentence below? In a democracy, everybody has the freedom to disagree with the government. English View Answer Need Expert Advice? Ask A Question Start Your Preparation with Prepp Mobile App Download the app from Google Play & App Store Download the app from Google Play & App Store Prepp Community Download Mobile App Government Exams & Jobs AFCAT CDS NDA UP Police UGC NET IAS Exam MPSC BPSC Railways RRB NTPC IBPS PO SBI CLERK SBI PO IBPS Clerk SSC CGL SSC CHSL CTET NTSE UPTET KVPY Previous Year Papers IAS Question Paper BPSC Question Paper IBPS PO Question Paper NDA Question Paper CDS Question Paper UGC NET Question Paper SSC CGL Question Paper RBI Grade B Question Paper RRB Question Paper SBI PO Question Paper Notes Current Affairs IAS Notes NDA Notes SBI PO Notes CDS Notes SSC CGL Notes SSC MTS Notes SSC CHSL Notes RRB ALP Notes RRB NTPC Notes About Us Contact Us Terms & Conditions Privacy Policy Prepp © 2025 Super Charge Your Preparation With Prepp+ Mock Test For All Government Exams Benefits of Prepp+ Subscription loading.. x
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Pfeifer & Jack Ferracane & Ronald L. Sakaguchi Master the use of dental materials with this all-in-one guide to restorative materials and procedures! Craig’s Restorative Dental Materials, Fifteenth Edition, addresses the fundamental concepts and skills needed to understand the science behin ...view more Master the use of dental materials with this all-in-one guide to restorative materials and procedures! Craig’s Restorative Dental Materials, Fifteenth Edition, addresses the fundamental concepts and skills needed to understand the science behind dental materials and their appropriate selection when designing and fabricating restorations. It begins with fundamentals and moves on to advanced skills in the manipulation of dental materials, providing insight on the latest advances and research along the way. From an expert author team, this comprehensive resource is considered to be the standard in the field of dental restorative materials. 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Offers are only available through our store and valid through October 15th with qualifying purchases. Terms and conditions apply. 5% OFF Skip to the end of the images gallery Subscription options Skip to the beginning of the images gallery Description Master the use of dental materials with this all-in-one guide to restorative materials and procedures! Craig’s Restorative Dental Materials,Fifteenth Edition,addresses the fundamental concepts and skills needed to understand the science behind dental materials and their appropriate selection when designing and fabricating restorations. It begins with fundamentals and moves on to advanced skills in the manipulation of dental materials, providing insight on the latest advances and research along the way. From an expert author team, this comprehensive resource is considered to be the standard in the field of dental restorative materials. New to this edition · NEW! Chapter on the principles of adhesion and adhesives· NEW! Photos highlighting the advances in digital technology in dentistry.· NEW! Short videos highlighting the key topics on each chapter.· UPDATED! Electronic resources including PPT files for instructors and board-exam-style clinical cases with discussion topics. Key Features NEW! Dedicated chapter covers the principles of adhesion and adhesives NEW!Current photos highlight the latest advances in digital technology in dentistry NEW!Enhancedebook version,included with every new print purchase, features key topics videos for each chapter and INBDE-style clinical cases, plus digital access to all the text, figures, and references, with the ability to search, customize content, make notes and highlights, and have content read aloud Comprehensive coverage ranges from fundamental concepts to advanced skills, detailing everything you need to know to select appropriate dental materials when designing and fabricating restorations Clear, focused approach provides an essential understanding of the fast-changing field of restorative dental materials More than 300 full-color illustrations show clinical detail with clarity and realism Logical organization arranges chapters by major clinical procedures Author Information By Carmem S. Pfeifer, DDS, PhD, Associate Professor Biomaterials and Biomechanics, OHSU, USA; Jack Ferracane, PhD, Department of Restorative Dentistry, School of Dentistry, Oregon Health and Science, University Portland, Oregon, USA and Ronald L. Sakaguchi, DDS, PhD, MS, MBA, Associate Dean for Technology and Innovation Professor, Biomaterials and Biomechanics, Department of Restorative Dentistry, Oregon Health and Science, University Portland, OR, USA Product Details More Information| ISBN Number | 9780323882767 | | Main Author | By Carmem S. Pfeifer, DDS, PhD, Jack Ferracane, PhD and Ronald L. Sakaguchi, DDS, PhD, MS, MBA | | Copyright Year | 2026 | | Edition Number | 15 | | Format | Book | | Trim | 216w x 276h (8.50" x 10.875") | | Imprint | Mosby | | Page Count | 262 | | Publication Date | 17 Apr 2025 | | Stock Status | IN STOCK | Table Of Contents 1. Role and Significance of Restorative Dental Materials Scope of Materials Covered in Restorative Dentistry A Systems Approach to Restorative Materials Application of Various Sciences Future Developments in Biomaterials 2. The Oral Environment Enamel The Mineral Dentin Physical and Mechanical Properties The Dentin–Enamel Junction Oral Biofilms and Oral Health Early Oral Biofilm Development on Enamel Oral Biofilm Maturation Oral Biofilm Development on Restorative and Implant Materials Interactions of Oral Biofilms With Common Restorative Materials Interactions of Oral Biofilms With Denture and Implant Materials Caries Prevention 3. Materials-Centered Treatment Design Evidence-Based Dentistry Patient Evidence Scientific Evidence Planning for Dental Treatment 4. Fundamentals of Materials Science Mechanical Properties Force Stress Stress-Strain Curves Viscoelasticity Dynamic Mechanical Properties Surface Mechanical Properties The Colloidal State Diffusion Through Membranes and Osmotic Pressure Adsorption, Absorption, and Sorption Surface Tension and Wetting Adhesion Optical Properties Color Measurement of Color Surface Finish and Thickness Opacity, Translucency, Transparency, and Opalescence Index of Refraction Optical Constants Thermal Properties Temperature Transition Temperatures Heat of Fusion Thermal Conductivity Specific Heat Thermal Diffusivity Coefficient of Thermal Expansion Electrical Properties Electrical Conductivity and Resistivity Dielectric Constant Electromotive Force Galvanism Electrochemical Corrosion Other Properties Tarnish and Discoloration Water Sorption Setting Time Shelf Life Biocompatibility and Tissue Reaction to Biomaterials Reactions of Pulp Reaction of Other Oral Soft Tissues to Restorative Materials Summary 5. General Classes of Biomaterials Metals and Alloys Chemical and Atomic Structure of Metals Atomic Structure Physical Properties of Metals Polymers Basic Nature of Polymers Principles of Polymerization Polymerization Mechanism Phases of Addition Polymerization Ceramics Composites 6. Preventive Dental Materials: Compounds Increasing Mineral Saturation and Diffusion Barriers Fluoride-Containing Agents and Materials Fluoride Varnishes Silver Diamine Fluoride Glass Ionomer to Prevent Progression of Caries Composition and Setting Resin-Modified Glass Ionomers Composition Properties Applications of GI and RMGI Casein Phosphopeptides-Amorphous Calcium Phosphate Composition Properties Application Clinical Studies Sodium Trimetaphosphate Composition Properties Applications Clinical Studies Bioactive Glass Composition Properties Applications Clinical Studies Calcium Hydroxide Composition Properties Applications Clinical Studies Tricalcium Silicate Cements Composition Properties Applications Clinical Studies Pit and Fissure Sealants Resin-Based Pit and Fissure Sealants Composition Properties Application of Resin-Based Sealants Flowable Composites as Sealants Clinical Outcomes Using Different Types of Sealants Resin Infiltration Composition Properties Applications Clinical Studies 7. Principles of Adhesion and Adhesive Systems Principles of Adhesion Types of Adhesion Adhesive Systems Bonding Substrates Bonding to Other Substrates Repair of Composite, Ceramic, and Ceramic-Metal Restorations 8. Polymeric Restorative Materials Resin Composites Multipurpose Resin Composites Composition Polymerization Reactions Packaging of Composites Properties of Composites Physical Properties Mechanical Properties Clinical Properties Composites for Special Applications Bulk-Fill Composites Flowable Composites Laboratory Composites Core Build-Up Composites Provisional Composites Light-Curing Units Quartz-Tungsten-Halogen Light-Curing Units Blue Light-Emitting Diodes Blue Light Hazards Prosthetic Applications of Polymers Physical Form and Composition Non-restorative Uses for Polymers: Athletic Mouth Protectors 9. Restorative Materials: Metals Metals for Direct Placement: Amalgam Composition and Morphology Amalgamation Processes: Admixed Alloys Physical and Mechanical Properties Bonding of Amalgam Dental Casting Alloys Types and Composition Metallic Elements Used in Dental Alloys Noble Alloys Base-Metal Alloys Wrought Alloys Microstructure Composition Properties Wrought Stainless Steel Alloys Wrought Nickel-Titanium Alloy Wrought Beta-Titanium Alloy 10. Restorative Materials: Ceramics Ceramics Classification of Dental Ceramics Classification by Application Classification by Fabrication Method Classification by Crystalline Phase General Applications of Ceramics in Prosthetic Dentistry Metal-Ceramic Crowns and Fixed Dental Prostheses All-Ceramic Inlays, Onlays, Veneers, Crowns, and Fixed Dental Prostheses Mechanical and Thermal Properties of Dental Ceramics Toughening Mechanisms Test Methods Comparative Data Optical Properties of Dental Ceramics All-Ceramic Restorations Sintered All-Ceramic Materials Heat-Pressed All-Ceramic Materials Machinable All-Ceramic Materials Metal-Ceramic Restorations Requirements for a Metal-Ceramic System Metal-Ceramic Bonding Ceramics for Metal-Ceramic Restorations Effect of Design on Metal-Ceramic Restorations Failure and Repair of Metal-Ceramic Restorations 11. Impression and Casting Material Clinical Use of Impression Materials Desirable Qualities of Impression Materials Types of Impression Materials Alginate Hydrocolloids Elastomeric Impression Materials Occlusal Registration Materials Impression Trays Cast or Die Materials Desirable Qualities of a Cast or Die Material Gypsum Products Chemical and Physical Nature of Gypsum Products Properties Manipulation Epoxy Die Materials Comparison of Impression and Die Materials Casting Investments Properties Required of an Investment Composition Calcium Sulfate-Bonded Investments Effect of Temperature on Investment Thermal and Hygroscopic Casting Investment Brazing Investment Investments for All-Ceramic Restorations 12. Cementation Classification and Characteristics of Luting Agents Classification Biocompatibility Interfacial Sealing and Anticariogenic Activity Adhesion Mechanical Properties Handling Properties and Radiopacity Viscosity and Film Thickness Solubility Aesthetics Acid-Base Cements Zinc Oxide–Eugenol and Non-eugenol Cements Glass Ionomer Resin-Modified Glass Ionomer Calcium Aluminate/Glass-Ionomer Cement Resin-Based Cements Resin Cements Self-Adhesive Resin Cements Resin Cements for Provisional Restorations 13. Technology Introduction Data Acquisition Digital Scan File Direct Scanning Indirect Scanning Computer-Aided Design Data Import Virtual Model Preparation Virtual Restoration Design Virtual Simulation Technologies Computer-Aided Manufacturing Subtractive Manufacturing Additive Manufacturing 14. Dental and Orofacial Implants Classification Endosseous Implant Osseointegration and Biointegration Factors Affecting the Endosteal Implant Geometry Magnitude of the Force Duration of the Force Type of Force Implant Diameter Implant Length Surfaces and Biocompatibility Ion Release Surfaces Surface Alterations Surface Coatings Implant Materials and Processing Challenges and the Future 15. Tissue Engineering Autograft Allograft Xenograft Alloplasts Strategies for Tissue Engineering Injection of Cells Guided Tissue Regeneration Cell Induction Cells Within Scaffold Matrices Stem Cells Biomaterials and Scaffolds Biological Materials Ceramic and Glass Materials Polymeric Materials Protein-Based Hydrogels Cell Culture Methods Tissue-Engineered Dental Tissues Appendix A: Conversion of Units Index Reviews Write Your Own Review Only registered users can write reviews. Please sign in or create an account product 322844 Craig's Restorative Dental Materials 115.89 121.99 USD InStock/Dentistry/Dental Materials/Complete Anatomy and Osmosis Offer Products/Product Format/Book 126 5936756 5936802 1 3 5 5936801 Master the use of dental materials with this all-in-one guide to restorative materials and procedures! Craig’s Restorative Dental Materials,Fifteenth Edition, addresses the fundamental concepts and skills needed to understand the science behind dental materials and their appropriate selection when designing and fabricating restorations. It begins with fundamentals and moves on to advanced skills in the manipulation of dental materials, providing insight on the latest advances and research along the way. From an expert author team, this comprehensive resource is considered to be the standard in the field of dental restorative materials. Master the use of dental materials with this all-in-one guide to restorative materials and procedures! Craig’s Restorative Dental Materials,Fifteenth Edition, addresses the fundamental concepts and skills needed to understand the science behind dental materials and their appropriate selection when designing and fabricating restorations. It begins with fundamentals and moves on to advanced skills in the manipulation of dental materials, providing insight on the latest advances and research along the way. From an expert author team, this comprehensive resource is considered to be the standard in the field of dental restorative materials.0 0 add-to-cart 9780323882767 2025 Professional By Carmem S. Pfeifer, DDS, PhD, Jack Ferracane, PhD and Ronald L. Sakaguchi, DDS, PhD, MS, MBA 2026 15 Book 216w x 276h (8.50" x 10.875")Mosby 262 Apr 17, 2025 IN STOCK By Carmem S. Pfeifer, DDS, PhD, Associate Professor Biomaterials and Biomechanics, OHSU, USA; Jack Ferracane, PhD, Department of Restorative Dentistry, School of Dentistry, Oregon Health and Science, University Portland, Oregon, USA and Ronald L. Sakaguchi, DDS, PhD, MS, MBA, Associate Dean for Technology and Innovation Professor, Biomaterials and Biomechanics, Department of Restorative Dentistry, Oregon Health and Science, University Portland, OR, USA Books, eBooks Book US No No No No Please Select Please Select No No Please Select Related Products Previous 7% OFF Book Netter's Concise Neuroanatomy Updated Edition Michael A. Rubin Oct 2016 Special Price$59.51$63.99 [x] Add to Cart Add to Wish ListAdd to Compare 7% OFF Flash Cards Netter's Advanced Head and Neck Flash Cards Neil S. 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https://www.themathdoctors.org/finding-the-radius-of-a-sphere/
Skip to content Finding the Radius of a Sphere June 20, 2018 January 11, 2024 / Algebra, AQOTW, Geometry, Trigonometry / Checking, Formulas / By Dave Peterson (An archive question of the week) An interesting question came to us in 2016, where rather than using a well-known formula, it was necessary to work out both what data to use, and how to calculate the desired radius. The question initially was vague: Compass, Ruler, and Radius — of a Sphere Using only a compass and a ruler, how can we calculate the radius of a hollow sphere? Obviously it is very easy, but I don't know how to even start! I guess we draw some random circles of equal radius on the sphere surface, but then what? I don't know what to do. I could think of many possible ways to do this, with various levels of accuracy, but I started with a mix of directness and humor: Hi, Kyriakos. I see nothing in the problem as stated that indicates it is "obviously very easy." What leads you to come to that conclusion? I imagine there are many methods you could pursue, depending on the precision desired, the kind of compass, and what you are willing to do with it. For starters, some compasses can be opened out enough to be able to use them as calipers: (I don't suppose you could just use the point of the compass to threaten bodily harm to the person who made the sphere if he doesn't tell you its radius. "Lateral thinking" puzzles work something like this, what with joke answers for finding the height of a skyscraper using only a barometer, or some other unlikely tool.) The reference to a barometer is discussed here. We had previously answered serious questions about finding the radius of a basketball (without restriction on tools), Finding the Radius of a Sphere and of a tennis ball, Radius of a Tennis Ball In the latter, I suggested a way to do it using three rulers. But clearly something more specific is in view here. I continued: My first serious thought involved using the compass to draw on the sphere: draw one circle on it with a known "radius" (a chord of the sphere); then put the compass point on that circle and draw another circle with the same "radius"; then use the compass to measure the straight-line distance between the two intersections. From that, you could in principle calculate the radius. But I don't want to work out a formula without knowing that's what I want to do. This turns out to be the method I will be using; but the initial question is open enough to keep thinking. I might be missing something really easy; or there might be more to the rules of the game. An easier way, though perhaps a little less exact in principle, is to measure the actual diameter of a drawn circle by putting the point on the circle and finding the greatest distance to a point on the other side of the circle. (This is something like the caliper idea, but doesn't require finding the exact opposite point on the sphere.) The "radius" is AB and the "diameter" is BD in the side view below: B /| / | / | /r |d/2 / | / | / | A-------C + | | | | | | | D From these measurements, there are several ways to calculate the radius R f the sphere, such as trigonometry or similar triangles. If I had continued with this thought, I would have added two radii to the figure and used the Pythagorean theorem: B /|\ / | \ / | \ /r |d/2 \R / | \ / | \ / | \ A-------C-----------O h | R-h | | | | | | D In triangle ABC, we have (defining c = d/2 to avoid fractions) (h^2 + c^2 = r^2), and in triangle OBC, ((R-h)^2 + c^2 = R^2). Solving the first for h and substituting in the second to eliminate h, we end up with (\displaystyle R = \frac{r^2}{2\sqrt{r^2 – c^2}}); replacing c with d/2 and simplifying, (\displaystyle R = \frac{r^2}{\sqrt{4r^2 – d^2}}). But we needed more information before pursuing any particular method. Pending more details from you, I think that leads to a decent answer, but I don't know that there isn't a much easier answer, perhaps even an "obvious" one. [One is to "rock" the ruler along the sphere to measure the arc length AB, then use that and r, but that requires numerical approximation rather than a formula.] So, can you tell me where the question came from? The reply added more rules, and a little context: Many thanks for your reply! To start with, I found this question on a Greek blog of math problems and riddles. I deduced it must be easy because several people had already claimed solutions. But they did not make these solutions visible, which only made my friends and me more curious. If you can work further on the first method, the one with the two intersecting circles, I would be grateful! For the other one, we must accept some approximation. “Easy”, of course, is in the mind of the beholder. And either method really requires approximation in practice; but the second does require a way to find opposite points on the circle, so although it might turn out to be just as accurate in practice, we are evidently looking for a theoretically precise answer. (I suspected there were also some hints as to the expected method, that I wasn’t privy to. Sometimes I will search for a problem across the web in hopes of finding the original wording, but this was in Greek so I couldn’t do that.) I said that, but Kyriakos reported explicitly this time that my first method was what he wanted: It seems that the two-circles method leads to a result that makes sense. Draw two overlapping circles of the same radius r. (Compared to the sphere radius, this has to be relatively small). Construct a segment to connect the two points where these intersect. Call its length "L," and measure this with the ruler. Then the sphere radius is (r/2)sqrt{(4r^2 - L^2)/((3r^2 - L^2)} But I cannot prove this result, so I would appreciate any assistance in explaining it. My friend, who gave me the original question, received this solution (or rather "reply") from someone else. It did not come with any explanation; and furthermore, we don't even know if it is correct (only by intuition). (Of course, you can’t really measure the segment directly with a ruler; he means to transfer the distance to a ruler using the compass.) So I dove into the harder but more interesting method, now seeking to derive the reported formula: I've been too busy to take the time to try to derive the formula until today, but this morning I managed it in perhaps ten minutes of free time. Then I had to write it up carefully.... Here is a picture: We have a sphere with radius R, center A, and two circles with radius r, centers B and C, that intersect at D and E. The distance DE is d. We want first to express d in terms of R and r, and then to solve for R. First consider isosceles triangle ABC, with legs R and base r. We conclude that the altitude of this triangle, from A to the midpoint of BC, is H = sqrt(R^2 - (r/2)^2) = sqrt(4R^2 - r^2)/2 For clarity, r is not really the radius of the circles (whose centers are inside the sphere), but the length of a chord from the “center” B or C to any point on the circumference. Likewise, DE is a straight line distance measured through the sphere. F is the midpoint of BC, and G is the midpoint of DE, both in the interior. Distance H is |AF| in the figure. I used GeoGebra to make the figure. In the next figure we have a cross-section through ADE; as seen here, the two circles look like the same ellipse, and B and C on the sphere coincide with F. But all the lines shown are coplanar. We first look at triangle ADF (which is not isosceles): Now consider the plane of triangle ADE: In triangle ADF, DF is the altitude of the equilateral triangle BCD, h = r sqrt(3)/2, and AF = H = sqrt(4R^2 - r^2)/2. By the law of cosines, cos(DAF) = [R^2 + H^2 - h^2]/[2RH] = [R^2 + (4R^2 - r^2)/4 - 3(r^2)/4]/[2R sqrt(4R^2 - r^2)/2] = [R^2 + R^2 - (r^2)/4 - 3(r^2)/4]/[R sqrt(4R^2 - r^2)] = [2R^2 - r^2]/[R sqrt(4R^2 - r^2)] Next, we look at triangle ADG, which shares the angle DAF, and then use the previously determined cosine to find an expression for d: In right triangle ADG, where G is the midpoint of DE, we have sin(DAF) = (d/2)/R Now, d = 2R sin(DAF) = 2R sqrt(1 - cos^2(DAF)) = 2R sqrt(1 - [2R^2 - r^2]^2/[R sqrt(4R^2 - r^2)]^2) = 2R sqrt(1 - [2R^2 - r^2]^2/[R^2 (4R^2 - r^2)]) = 2R sqrt(R^2 (4R^2 - r^2) - [2R^2 - r^2]^2]/[R^2 (4R^2 - r^2)]) = 2R sqrt([4R^4 - r^2R^2 - 4R^4 + 4r^2R^2 - r^4]/[4R^4 - r^2R^2]) = 2R sqrt([3r^2R^2 - r^4]/[4R^4 - r^2R^2]) = 2 sqrt([3r^2R^2 - r^4]/[4R^2 - r^2]) We have now expressed d in terms of r and R; now we solve for R, as planned: Solving this for R, d^2 = 4[3r^2R^2 - r^4]/[4R^2 - r^2] d^2 [4R^2 - r^2] = 4[3r^2R^2 - r^4] 4d^2R^2 - d^2r^2 = 12r^2R^2 - 4r^4 4d^2R^2 - 12r^2R^2 = d^2r^2 - 4r^4 [4d^2 - 12r^2]R^2 = d^2r^2 - 4r^4 R^2 = [d^2r^2 - 4r^4]/[4d^2 - 12r^2] = r^2[d^2 - 4r^2]/[4(d^2 - 3r^2)] R = r/2 sqrt([d^2 - 4r^2]/[d^2 - 3r^2]) = r/2 sqrt([4r^2 - d^2]/[3r^2 - d^2]) And that's your formula. Formatted nicely, the formula is (\displaystyle R = \frac{r}{2} \sqrt{\frac{4r^2 – d^2}{3r^2 – d^2}}). I’ve played with this and see no nice, “obvious” geometrical meaning for it. But one thing we can do to check it out is to try some special cases. First, what if the sphere is really a plane (that is, the radius in “infinite”)? Then we just have the familiar figure of two overlapping circles in a plane, and can easily calculate that (d = r\sqrt{3}) (twice the altitude of an equilateral triangle). Putting that into the formula, (\displaystyle R = \frac{r}{2} \sqrt{\frac{4r^2 – 3r^2}{3r^2 – 3r^2}} = \frac{r}{2} \sqrt{\frac{r^2}{0}}) which is undefined as expected. Second, what if each circle were a great circle? Then (r = R\sqrt{2}) and (d = 2R). Putting those into the formula, we get (\displaystyle R = \frac{R\sqrt{2}}{2} \sqrt{\frac{8R^2 – 4R^2}{6R^2 – 4R^2}} = \frac{R\sqrt{2}}{2} \sqrt{\frac{4R^2}{2R^2}} = \frac{R\sqrt{2}}{2} \sqrt{2} = R) as expected. 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7488
https://www.cuemath.com/ncert-solutions/ncert-solutions-class-10-maths-chapter-10-circles/
NCERT Solutions Class 10 Maths Chapter 10 Circles - Access free PDF We use cookies on this site to enhance your experience. To learn more, visit our Privacy Policy OK Grade KG 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th 11th 12th Online Training About Us Already booked a tutor? Sign up Grade KG 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th 11th 12th Online Training About Us NCERT Solutions Class 10 Maths Chapter 10 Circles NCERT Solutions for Class 10 Maths Chapter 10 Circles is majorly focused on giving students a deep-seated understanding of tangents and their properties. A circle can be defined as a closed plane geometric shape where the locus of a point moves around a fixed point at a fixed gap such that the outline is equidistant from the center point. The chapter starts by elaborating on the position of a line with respect to a circle and how they are related. This line can be non-intersecting, meet the circle at two points meaning that it lies within the circle (secant), and touch the circle at one point only (tangent). We can also say that the tangent to a circle is a special case of the secant when the two endpoints of its corresponding chord coincide. This lesson consists of a host of theorems that make solving questions easier. Additionally, kids have the opportunity to learn about various terminologies associated with a tangent. As circles form an integral part of geometry, therefore, kids must go through the NCERT solutions class 10 Maths Chapter 10 with care and laser concentration. This lesson is of great importance as the underlying concepts are used in several sister topics such as areas related to circles, constructions, and coordinate geometry. The Class 10 Maths NCERT Solutions Chapter 10 provides a wide variety of problem sums that can challenge a kid’s understanding eventually instilling ironclad concepts within him. With the help of this article, we will do a detailed analysis of NCERT Solutions Chapter 10 Circles and also you can find some of these in the exercises given below. NCERT Solutions Class 10 Maths Chapter 10 Ex 10.1 NCERT Solutions Class 10 Maths Chapter 10 Ex 10.2 NCERT Solutions for Class 10 Maths Chapter 10 PDF NCERT solutions Class 10 maths chapter 10 circles make use of various activities to teach children new concepts. It provides fun ways to get a practical demonstration of the crux of the topic. For example, to draw water from a well a pulley is required with a rope on it. Here this rope, if considered as a ray, is like a tangent to the circle representing the pulley. Thus, using real-life examples these solutions help students explore concepts in an elaborate manner. The free PDF of NCERT Solutions Class 10 Maths Chapter 10 circles is given below. ☛ Download Class 10 Maths NCERT Solutions Chapter 10 Circles NCERT Class 10 Maths Chapter 10Download PDF NCERT Solutions for Class 10 Maths Chapter 10 Circles Chapter 10 helps students understand the concepts related to Circles and their properties. This chapter is an important section of geometry and questions related to it are asked not just in board exams but also competitive exams. With the help of these solutions, students can learn the concept of tangents to a circle and discover the relationship between the radius as well as the tangent. As these concepts will make their way over and again throughout the next chapters it is highly recommended to periodically revise them. Now, let us go through the detailed analysis of NCERT Solutions Class 10 Maths Chapter 10 Circles Class 10 Maths Chapter 10 Exercise 10.1 - 4 questions Class 10 Maths Chapter 10 Exercise 10.2 - 13 questions ☛ Download Class 10 Maths Chapter 10 NCERT Book Topics Covered:Class 10 Maths NCERT Solutions Chapter 10 Circles cover a variety of topics such as tangents, length of tangents, normal to a circle, radius-tangent relationship, etc. Questions related to these topics are solved in a step-wise manner using the properties based on the aforementioned topics. Total Questions: Class 10 Maths Chapter 10 Circles consists of a total of 17 sums, of which 10 are easy, 3 are moderately complex, and 4 are difficult problems. List of Formulas in NCERT Solutions Class 10 Maths Chapter 10 NCERT solutions Class 10 Maths Chapter 10 covers key points that are important for exams. This chapter introduces students to some new terms like the point of contact, length of tangent, and tangent to a circle. In order to solve questions related to this exercise, we need to use some formulas from the previous classes. In addition to this, there are many theorems that are equally important. To get a holistic understanding of this lesson children must combine the topics learned in earlier classes with the newer subject matter. These theorems and formulas can be seen as tools to simplify a complicated question. Let us do a quick overview of the formulas which are frequently used in NCERT Solutions Class 10 Maths Chapter 10: Area of circle = πr 2 Circumference of a circle= 2πr Arc length of the sector of angle A= (A/360) 2πr There is only one tangent at a point of the circle. The common point of the tangent and the circle is called the point of contact. The line containing the radius through the point of contact is also sometimes called the ‘normal’ to the circle at the point Important Questions for Class 10 Maths NCERT Solutions Chapter 10 | CBSE Important Questions for Class 10 Maths Chapter 10 Exercise 10.1 | | Question 1 Question 2 Question 3 Question 4 | | CBSE Important Questions for Class 10 Maths Chapter 10 Exercise 10.2 | | Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 Question 11 Question 12 Question 13 | Video Solutions for Class 10 Maths NCERT Chapter 10 | NCERT Video Solutions for Class 10 Maths Chapter 10 | | Video Solutions for Class 10 Maths Exercise 10.1 | | Question 1 | Question 2 | | Question 3 | Question 4 | | Video Solutions for Class 10 Maths Exercise 10.2 | | Question 1 | Question 2 | | Question 3 | Question 4 | | Question 5 | Question 6 | | Question 7 | Question 8 | | Question 9 | Question 10 | | Question 11 | Question 12 | | Question 13 | | FAQs on NCERT Solutions Class 10 Maths Chapter 10 Why are NCERT Solutions Class 10 Maths Chapter 10 Important? NCERT Solutions for Class 10 Maths Chapter 10 help students understand the basics of geometry. Kids can use this chapter to explore simple and advanced concepts of a circle. Moreover, students can also revise topics from the previous classes. Thus, these solutions are important as it helps children lay a strong foundation for higher classes. Additionally, it also shows kids how to present their solutions in an examination to get excellent marks. Do I Need to Practice all Questions Provided in NCERT Solutions Class 10 Maths Circles? All questions in NCERT Solutions Class 10 Maths Circles are based on important concepts that give a 360 degree view of the lesson. As the problems can prove to be complex, practicing all the sums along with the examples can help students master the topic of circles and understand the application of various formulas. Since the sums are proof-based, kids must also revise the theorems and practice questions twice. What are the Important Topics Covered in NCERT Solutions Class 10 Maths Chapter 10? All topics in the NCERT Solutions Class 10 Maths Chapter 10 Circles are equally important. Thus, it is advised to give an ample amount of time to all the sections. Learning the theorems and understanding their proofs is also essential to attempt the problems. There are two theorems in this lesson on which all the sums are based as well as certain inferences have been drawn that form the pillars of Circles. How Many Questions are there in NCERT Solutions Class 10 Maths Chapter 10 Circles? The total number of questions in NCERT Solutions Class 10 Maths Chapter 10 Circles is 17 that have been divided among 2 exercises. These sums are curated in different formats in order to make learning engaging and interesting. Some of these problems are super easy, while others require a little brainstorming and revision. What are the important formulas in Class 10 Maths NCERT Chapter 10 Circles? NCERT Solutions Class 10 Maths Chapter 10 is more driven towards theorems rather than formulas. Thus, kids need to focus on understanding these theories and also review certain concepts from previous classes. Some of the complicated sums make use of the properties of other polygons. One needs to solve a variety of questions based on these in order to develop a rock-solid foundation. How can CBSE Students utilize NCERT Solutions Class 10 Maths Chapter 10 effectively? In order to utilize the NCERT Solutions Class 10 Maths Chapter 10, students need to solve each and every question from both the exercises. In case they get stuck in any of these problems, they can refer to the formulas and theory. Kids often tend to ignore the ‘solved examples’ of the chapter, but attempting these gives them an added value. Additionally, it is important to first get well-versed with the theory before moving further. NCERT Solutions Class 10 Maths Chapters Chapter 1 Real NumbersChapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two VariablesChapter 4 Quadratic Equations Chapter 5 Arithmetic ProgressionsChapter 6 Triangles Chapter 7 Coordinate GeometryChapter 8 Introduction to Trigonometry Chapter 9 Some Applications Of TrigonometryChapter 11 Constructions Chapter 12 Areas Related to CirclesChapter 13 Surface Areas and Volumes Chapter 14 StatisticsChapter 15 Probability Explore math program Math worksheets and visual curriculum Sign up FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATHS PROGRAM Maths Program Home Tuitions Home Tutors Private Tutors Tuition Tutors Online Tuitions Online Maths Courses Math Tutors CURRICULUM 1st Grade Maths 2nd Grade Maths 3rd Grade Maths 4th Grade Maths 5th Grade Maths 6th Grade Maths 7th Grade Maths 8th Grade Maths ABOUT US Our Mission Our Journey Our Team MATH TOPICS Numbers Algebra Geometry Measurement Commercial Maths Data Trigonometry Calculus Maths Formulas Calculators Multiplication Tables NCERT Solutions NCERT Solutions for Grade 7 NCERT Solutions for Grade 8 NCERT Solutions for Grade 9 NCERT Solutions for Grade 10 QUICK LINKS Maths Games Maths Puzzles Maths Questions Blogs Maths Olympiad (IMO) FAQs INTERNATIONAL United States United Kingdom Canada Dubai Australia France Germany Indonesia Italy Netherlands Sri Lanka Singapore Saudi Arabia Oman Bahrain Qatar Norway Sweden Ireland MATHS WORKSHEETS Kindergarten Maths Worksheets Maths Worksheet for Grade 1 Maths Worksheet for Grade 2 Maths Worksheet for Grade 3 Maths Worksheet for Grade 4 Maths Worksheet for Grade 5 Maths Worksheet for Grade 6 Maths Worksheet for Grade 7 Maths Worksheet for Grade 8 Maths Worksheet for Grade 9 Maths Worksheet for Grade 10 FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATHS PROGRAM Maths Program Home Tuitions Home Tutors Private Tutors Tuition Tutors Online Tuitions Online Maths Courses Math Tutors MATHS WORKSHEETS Kindergarten Maths Worksheets Maths Worksheet for Grade 1 Maths Worksheet for Grade 2 Maths Worksheet for Grade 3 Maths Worksheet for Grade 4 Maths Worksheet for Grade 5 Maths Worksheet for Grade 6 Maths Worksheet for Grade 7 Maths Worksheet for Grade 8 Maths Worksheet for Grade 9 Maths Worksheet for Grade 10 INTERNATIONAL United States United Kingdom Canada Dubai Australia France Germany Indonesia Italy Netherlands Sri Lanka Singapore Saudi Arabia Oman Bahrain Qatar Norway Sweden Ireland ABOUT US Our Mission Our Journey Our Team MATH TOPICS Numbers Algebra Geometry Measurement Commercial Maths Data Trigonometry Calculus Maths Formulas Calculators Multiplication Tables NCERT Solutions NCERT Solutions for Grade 7 NCERT Solutions for Grade 8 NCERT Solutions for Grade 9 NCERT Solutions for Grade 10 QUICK LINKS Maths Games Maths Puzzles Maths Questions Blogs Maths Olympiad (IMO) FAQs CURRICULUM 1st Grade Maths 2nd Grade Maths 3rd Grade Maths 4th Grade Maths 5th Grade Maths 6th Grade Maths 7th Grade Maths 8th Grade Maths Terms and ConditionsPrivacy Policy
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https://www.thoughtco.com/hierarchy-syntax-term-1690835
Skip to content Humanities › English › English Grammar › Hierarchy in Grammar English English Grammar An Introduction to Punctuation Writing By Richard Nordquist Richard Nordquist English and Rhetoric Professor Ph.D., Rhetoric and English, University of Georgia M.A., Modern English and American Literature, University of Leicester B.A., English, State University of New York Dr. Richard Nordquist is professor emeritus of rhetoric and English at Georgia Southern University and the author of several university-level grammar and composition textbooks. Learn about our Editorial Process Updated on February 12, 2020 In grammar, hierarchy refers to any ordering of units or levels on a scale of size, abstraction, or subordination. Adjective: hierarchical. Also called syntactic hierarchy ormorpho-syntactic hierarchy. The hierarchy of units (from smallest to largest) is conventionally identified as follows: Phoneme Morpheme Word Phrase Clause Sentence Text Etymology:From the Greek, "rule of the high priest" Examples and Observations Charles Barber, Joan C. Beal, and Philip A. Shaw: Within the sentence itself, there is a hierarchical structure. Take a simple sentence: (a) The women were wearing white clothes. This can be divided into two parts, Subject and Predicate, in each of which there is a main part and a subordinate part. The Subject consists of a Noun Phrase ('The women'), in which a noun ('women') is the head, and a determiner ('The') is a modifier. The Predicate has as its head a Verb Phrase ('were wearing') which governs a Noun Phrase ('white clothes') as its Object. The Verb Phrase has a main verb ('wear') + -ing as its head, and an auxiliary ('were') as a subordinate part, while the Noun Phrase has as its head a noun ('clothes'), and an adjective ('white') as a modifier... This notion of hierarchy in sentence structure is of primary importance. For example, if we wish to change a sentence (for example, from a statement to a question, or from an affirmative to a negative form), we cannot do it by rules which just shuffle individual words around: the rules have to recognize the various units of the sentence and the ways in which they are subordinated to one another. For instance, if we want to turn the sentence 'The king is at home' into a question, we have to bring 'is' in front of the whole noun phrase 'the king' to produce 'Is the king at home?' "The is king at home?" would be ungrammatical. C.B. McCully: Turning to a syntactic hierarchy, we might want to observe that the smallest elements of syntax are morphemes. Whether these morphemes are either nonlexical (as in the plural inflections /s/ or /iz/ -- cats, houses) or lexical (= lexeme -- cat, house), their function is to constitute words; words are gathered into syntactic phrases; phrases are gathered into sentences . . . and beyond the sentence, if we wish our hierarchical theory to account for reading as well as speaking and writing, we could include constituents such as the paragraph. But clearly, morpheme, word, phrase and sentence are again constituents of the syntactic grammar of English. Charles E. Wright and Barbara Landau: The relationship between semantic and syntactic levels has been actively debated (see, e.g., Foley & van Valin, 1984; Grimshaw, 1990; Jackendoff, 1990). However, one general framework posits linking rules, building on the fact that the semantic and syntactic levels of representation share a similar hierarchical structure: Those thematic roles highest in the thematic hierarchy will be assigned to those structural positions highest in the syntactic hierarchy. For example, in the thematic hierarchy, the role of agent is considered to be 'higher' that either 'patient' or 'theme'; in the grammatical hierarchy, the syntactic function of subject is assumed to be higher than direct object, which is higher than indirect object (see, e.g., Baker, 1988; Grimshaw, 1990; Jackendoff, 1990). Aligning these two hierarchies will have the net result that, if there is an agent to be expressed in the sentence (e.g., using the verb give), that role will be assigned to subject position, with the patient or theme assigned to direct object. Marina Nespor, Maria Teresa Guasti, and Anne Christophe: In prosodic phonology, it is assumed that, besides a syntactic hierarchy, there is a prosodic hierarchy. The former is concerned with the organization of a sentence into syntactic constituents and the latter with the analysis of a string into phonological constituents. The prosodic hierarchy is built on the basis of the morpho-syntactic hierarchy. Although there is a reliable correlation between the two hierarchies, the correlation is not always perfect (cf. also Chomsky and Halle 1968). A classical example of the mismatch between syntax and prosody is illustrated below: (12) [This is ]]] (13) [This is the dog] [that chased the cat] [that bit the rat] [that . . . In (12), the bracketing indicates the relevant syntactic constituents, specifically NP's. These constituents do not correspond to the constituents of the prosodic structure of the sentence, which are indicated in (13). Format mla apa chicago Your Citation Nordquist, Richard. "Hierarchy in Grammar." ThoughtCo, Apr. 5, 2023, thoughtco.com/hierarchy-syntax-term-1690835. Nordquist, Richard. (2023, April 5). Hierarchy in Grammar. Retrieved from Nordquist, Richard. "Hierarchy in Grammar." ThoughtCo. (accessed July 24, 2025). listeme (words) Binomials in English: Definition and Examples Elsewhere Principle in Linguistics Understanding Anacoluthon (Syntactic Blend) in English Grammar Head (Words) Colligation Enallage Late Closure (Sentence Processing) sluicing (grammar) Diglossia in Sociolinguistics Syntactic Ambiguity Coordinating Conjunctions in English Definition and Examples of Transformations in Grammar lexeme (words) Definition, Etymology and Examples Definition and Examples of Correctness in Language Definition of Deep Structure By clicking “Accept All Cookies”, you agree to the storing of cookies on your device to enhance site navigation, analyze site usage, and assist in our marketing efforts.
7490
https://pmc.ncbi.nlm.nih.gov/articles/PMC4815482/
Intermittent Diazepam versus Continuous Phenobarbital to Prevent Recurrence of Febrile Seizures: A Randomized Controlled Trial - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer | PMC Copyright Notice Iran J Child Neurol . 2016 Winter;10(1):21–24. Search in PMC Search in PubMed View in NLM Catalog Add to search Intermittent Diazepam versus Continuous Phenobarbital to Prevent Recurrence of Febrile Seizures: A Randomized Controlled Trial Mohammadreza SALEHIOMRAN Mohammadreza SALEHIOMRAN, MD 1 Pediatric Neurologist, NonCommunicable Pediatric Diseases Research Center, Babol University of Medical Sciences, Babol, Iran Find articles by Mohammadreza SALEHIOMRAN 1, Seyed Mohammad HOSEINI Seyed Mohammad HOSEINI, MD 2 Pediatrician, Non-Communicable Pediatric Diseases Research Center, Babol University of Medical Sciences, Babol, Iran Find articles by Seyed Mohammad HOSEINI 2, Ali GHABELI JUIBARY Ali GHABELI JUIBARY, MD 3 Department of Neurology, Mashhad University of Medical Sciences, Mashhad, Iran Find articles by Ali GHABELI JUIBARY 3 Author information Article notes Copyright and License information 1 Pediatric Neurologist, NonCommunicable Pediatric Diseases Research Center, Babol University of Medical Sciences, Babol, Iran 2 Pediatrician, Non-Communicable Pediatric Diseases Research Center, Babol University of Medical Sciences, Babol, Iran 3 Department of Neurology, Mashhad University of Medical Sciences, Mashhad, Iran ✉ Corresponding Author: Salehiomran MR. MD, Non Communicable Pediatric Diseases Research Center, Babol University of Medical Sciences, Babol, Iran , Phone Number: +989111144527 , Email: salehiomran@yahoo.com Received 2014 Feb 22; Revised 2014 Jun 29; Accepted 2014 Jul 9. This is an Open Access article distributed under the terms of the Creative Commons Attribution License, ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC4815482 PMID: 27057183 Abstract Objective Febrile seizure is the most common neurologic problem in children between 3 months to 5 years old. Two to five percent of children aged less than five yr old will experience it at least one time. This type of seizure is age dependent and its recurrence rate is about 33% overalls and 50% in children less than one yr old. The prophylactic treatment is still controversial, so we conducted a randomized controlled clinical trial to find out the effectiveness of continuous phenobarbital versus intermittent diazepam for febrile seizure. Materials & Methods This clinical trial was conducted in the Department of Pediatric Neurology, Babol University of Medical Sciences, Babol, Iran between March 2008 and October 2010. All children from 6 month to 5 yr old referred to Amirkola Children’s Hospital, Babol, Iran were enrolled in the study. Children with febrile seizure that had indication for prophylaxis but did not receive any prophylaxis previously were enrolled in the study. For prophylactic anti convulsion therapy, patients were divided randomly in two groups. One group received continuous phenobarbital and another treated with intermittent diazepam whenever the children experienced an episode of febrile illness for up to one year after their last convulsion. Results Of all 145 studied cases, the recurrent rate in children under prophylaxis with diazepam was 11/71 and in phenobarbital group was 17/74. There was no significant difference in the recurrence rate in both groups. Conclusion There was no significant difference in the effectiveness of phenobarbital and diazepam in prevention of recurrent in febrile seizure and we think that in respect of lower complication rate in diazepam administration, it cloud be better choice than phenobarbital. Key Words: Prophylaxis, Febrile Seizures, Diazepam, Phenobarbital Introduction Febrile Seizure (FC) is one of the most common transient neurologic disorders in infants and childhood. Its prevalence is about 2-5% and the mean age at onset is 3 month to 5 yr old (1). Even the prognosis is good for the most patients, but seizures are upsetting to the children and parents. Prevention of the next seizures is important for prevention of brain hypoxic damage and sudden fall related injuries (2, 3). Fortunately, in most cases, there is no need to treatment but in some cases with high risk for recurrence prophylaxis must be considered. The overall rate of recurrence after the first episode is about 30%-37% but in child less than 1 year old, it could be about 50% (4). Some recurrence risk factors include age less than one yr old, the history of febrile seizure in fist – degree relatives of seizures occur at temperatures less than 39°C and when there is low Interval between fever and seizure (5). In child with one episode of recurrence, the risk of subsequent recurrence increases and 9-17% of them will experience more than 3 times recurrence. Three quarters of it occurs within the first year after seizure (5). Intermittent prophylaxis with oral diazepam during the first 3 days of febrile illnesses is associated with less complication such as lethargy and ataxia (6) but permanent prophylaxis with oral phenobarbital may be associated with complications like behavioral disturbances, irritability, hyperactivity and decreased of cognitive function (7). Use of rectal diazepam during an episode of seizure is another method of prophylaxis not often considered because of its failure to prevent the onset of seizure and its age related limitation. It is used to prevent seizure in 3 to 5 yr-old children and thus prevent prolongation of seizures and its recurrence until next 12 hours (6, 8). To evaluate the efficacy of different anticonvulsive regimes for preventing febrile seizures, many randomized clinical trials muse be done. In this study, we conducted a single blind, randomized clinical trial to compare effectiveness of intermittent oral diazepam versus continuous phenobarbital in children with febrile convolution. Materials & Methods This single blind randomized clinical trial, registered as RCT of code No. CT2015080223393N2, was done in the Department of Pediatric Neurology, Babol University of Medical Sciences, Babol, Iran between March 2008 and October 2010. All children 6 months to 5 yr old with recurrent simple FC(≥ 3 episode) or complex FC that did not receive any anticonvulsive drug ,were enrolled into the study. Simple FC were defined as brief (< 15 minutes), generalized seizure in association with fever and only once during a 24-hour period. Childe with history of neonatal seizure, seizure without fever, chronic disease related electrolytes imbalance and history of anticonvulsive consumption, were excluded. The study was approved by the regional Ethics Committee of Babol University of Medical Sciences. After getting informed consent of child parents, all of them randomly were divided in tow prophylaxis groups. One group received oral phenobarbital (Tab 15 and 60 mg), 3-5 mg/kg/day in tow divided doses for at least one year and another group was recommended to use oral diazepam 0.33 mg/kg/TDS during febrile illness for two days without antipyretics consumption. We recommended that the child should be admitted to hospital if new convulsions occurred. Basic data on the mother’s pregnancy, the child’s birth, neonatal period, sex, age, previous febrile convulsions with detailed description of first convulsion, and epilepsy in parents or siblings were obtained from medical document and parents. All cases were fallowed regularly up to one year after their convulsion. In each monthly call, parents were questioned about occurrence of fever, new febrile convulsions, compliance with instructions, drug related side effects, and assesses the child’s clinical progress, including possible side effects of medication. In case of unable to follow the patient, or other diagnoses achieved, he/she was excluded. The main outcome was report of febrile seizures by the parents. Data were analyzed using the chi-squared test or Fisher’s exact test using the SPSS version 21 (Chicago, IL, USA). P less than 0.05 were considered significant. Results Of all 154 studied patients, nine patients excluded because of the aforementioned causes. Out of the remaining 145 patients, 71 were in the diazepam and 74 in the phenobarbital group. The mean age of them was 22.61±9.11 and 20.59±7.93 months respectively (P =0.158). There was no significant difference between the mean age of cases in both groups and according to the Fig 1, the age of patient followed normal distribution curve. Table 1 shows the demographic data of patient in both groups and according to Fishers Exact Test result, there was no significant difference between the base line variable in both groups. In one year’s fallow up of patients, we had recurrence in 11 (15.5%) cases of diazepam group and in 17 cases (23%) of phenobarbital group. There was no significant difference between the recurrence rate in both groups (P=0.296). Side effects of phenobarbital like hyperkinesia, irritability, and restlessness were observed in some patients but diazepam related side effects except sedation were not seen. Fig 1. Open in a new tab The age of patient follows normal distribution curve Table1. Clinical Data Phenobarbital versus Diazepam | Group | Phenobarbital n (%) | Diazepam n (%) | Fishers Exact Test | :---: :---: | | Male | 46 (62.2) | 41(57.7) | 0.614 | | Female | 28 (37.8) | 30(42.3) | | Recurrent Febrile Seizure | 50 (67.6) | 50(70.4) | 0.724 | | Complex Febrile Seizure | 24 (32.4) | 21(29.6) | Open in a new tab Discussion We did not find any significant difference of two studied prophylaxis protocols but the recurrence rate by diazepam was less than phenobarbital (15.5% vs. 23%). Most previous studies have compared two drugs with placebo and clinical trials comparing these together are fewer. In Italy, the rate of febrile seizure recurrence with oral diazepam was about 11.1% compared to the 30.7% recurrence rate in those who did not receive any treatment, was substantially reduced. Lack of timely detection of fever by parents was one of the diazepam administration problems in addition to ataxia and lethargy as side effects (9). Rose and colleagues (11) conducted a study on clobazam and the ataxia rate was less than diazepam. In a study, recurrent febrile seizure risk, significantly reduced with intermittent oral diazepam versus placebo or phenobarbital versus placebo (10). However, parents and families should be supported with adequate contact details of medical services and information on recurrence, first aid management and, most importantly, the benign nature of the phenomenon (10). Efficacy of diazepam in preventing febrile seizure compared with placebo is obvious, but it was not statistically significant (2, 3). In conclusion, Diazepam is a safe and quickly absorbed anticonvulsant that its peak serum level in oral consumption, reach more rapidly in children and we think in respect of the well-known side effects of phenobarbital, it can be a good choice for prophylaxis of febrile seizure. Acknowledgements The authors wish to thank the Amirkola Children’s Hospital Clinical Research Development Committee especially Dr Mohaddese Mirzapour and Faeze Aghajanpour for their contribution to this study and all parents and their children who participated in this study. Conflict of Interest: This study was supported by a research grant and Residency thesis of Dr Seyed Mohammad Hoseini from the Non-Communicable Pediatric Diseases Research Center of Babol University of Medical Sciences (Grant Number: 76144). Authors’ Contribution: DR Salehiomran: Development of original idea, study concept, and writing the manuscript Dr Hoseini: Study concept and design, collecting of data Dr Ghabeli Juibary: Study concept and design, analysis of data References 1.Nelson KB, Ellenberg JH. Prognosis in children with febrile seizures. Pediatrics. 1978;61:720–727. [PubMed] [Google Scholar] 2.Rosman NP, Colton T, Labazzo J, Gilbert PL, Gardella NB, Kaye EM, et al. A controlled trial of diazepam administered during febrile illnesses to prevent recurrence of febrile seizures. N Engl J Med. 1993;329:79–84. doi: 10.1056/NEJM199307083290202. [DOI] [PubMed] [Google Scholar] 3.Autret E, Ployet JL. Traitement des convulsions fébriles. Arch Pédiatr. 2002;9:91–95. [PubMed] [Google Scholar] 4.Annegers JF, Hauser WA, Shirts SB, et al. Factors prognostic of unprovoked seizure after febrile convulsion. N Engl J Med. 1987;316:493–498. doi: 10.1056/NEJM198702263160901. [DOI] [PubMed] [Google Scholar] 5.Sulzbacher S, Farwell JR, Temkin N, Lu AS, Hirtz DG. Late cognitive effects of early treatment with phenobarbital. Clin Pediatr. 1999;38:387–394. doi: 10.1177/000992289903800702. [DOI] [PubMed] [Google Scholar] 6.Millichap JG, Colliver JA. Management of febrile seizures: survey of current practice and phenobarbital usage. Pediatr Neurol. 1991;7:243–248. doi: 10.1016/0887-8994(91)90039-n. [DOI] [PubMed] [Google Scholar] 7.Thilothammal N, Krishnamurthy PV, Kamala KG, Ahamed S, Banu K. Role of phenobarbitone in preventing recurrence of febrile convulsions. Indian Pediatr. 1993;30:637–642. [PubMed] [Google Scholar] 8.Farwell JR, Lee YJ, Hirtz DG, Sulzbacher SI, Ellenberg JH, Nelson KB. Phenobarbital for fibril seizures: effects on intelligence and on seizure recurrence. N Engl J Med. 1990;322:364–369. doi: 10.1056/NEJM199002083220604. [DOI] [PubMed] [Google Scholar] 9.Verroti A, Latini G, Trotta D, et al. Intermittent oral DZP prophylaxis in FC: its effectiveness for FC. Eur J Ped Neurol. 2004;8:131–4. doi: 10.1016/j.ejpn.2004.01.008. [DOI] [PubMed] [Google Scholar] 10.Offringa M, Newton R. Prophylactic drug management for febrile seizures in children. Cochrane Database Syst Rev. 2012;18(4):CD003031. doi: 10.1002/14651858.CD003031.pub2. doi: 10.1002/14651858. CD003031. pub2. [DOI] [PubMed] [Google Scholar] 11.Rose W1, Kirubakaran C, Scott JX. Intermittent Clobazam therapy in FC. Indian J Pediatr. 2005 Jan;72(1):31–3. doi: 10.1007/BF02760577. [DOI] [PubMed] [Google Scholar] Articles from Iranian Journal of Child Neurology are provided here courtesy of Shahid Beheshti University of Medical Sciences ACTIONS PDF (312.5 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Materials & Methods Results Discussion Acknowledgements Authors’ Contribution: References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
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https://www.popai.pro/resources/homework/given-the-unit-circle-find-the-length-of-the-chord-subtending-an-angle-of-%CE%B8-at-the-center-explain-the-steps-and-provide-the-solution/
Given the unit circle, find the length of the chord subtending an angle of θ at the center Explain the steps and provide the solution - PopAi AI Presentation Templates AI ChatPDF AI Writing AI Paraphraser AI Image Resources Pricing Login Sign up Login Try for free AI PresentationAI Image & VideoAI ChatPDFAI Writing Reports Work Report Business Report Academic Report Literary Analysis Financial Report Analysis Meeting Notes Analysis Project Report Analysis Quotes Quotes for Myself Quotes for Students Quotes about Life Quotes about Wisdom Quotes about Reading Quotes of the Day Quotes about Teamwork Love Quotes Essays Essay Topics Essay Guides Essay Sample Celebration Celebration Dinner Celebration Ceremony Messages Labor Day Birthday Party Birthday Wishes Love Letters Love Messages Good Morning Wishes New Year Wishes Homework Humanities Math Science Email Resumes Speeches Teaching Tech Home>Resources>Homework>Math Given the unit circle, find the length of the chord subtending an angle of $ heta$ at the center. Explain the steps and provide the solution. Answer 1 Ava Martin To find the length of the chord subtending an angle $\theta$ at the center of the unit circle, we can use the formula for the chord length: $L = 2r \sin\left(\frac{\theta}{2}\right)$ Since the radius $r$ of the unit circle is 1, the formula simplifies to: $L = 2 \sin\left(\frac{\theta}{2}\right)$ Let’s go through the steps: Consider the unit circle with the center at the origin (0,0) and radius 1. The chord subtends an angle $\theta$ at the center. Draw radii from the center to the endpoints of the chord, forming an isosceles triangle. Drop a perpendicular from the center to the chord, bisecting the angle $\theta$ and the chord. The length of the half-chord is $\sin\left(\frac{\theta}{2}\right)$, thus the full chord length is: $L = 2 \sin\left(\frac{\theta}{2}\right)$ Therefore, the length of the chord is: $L = 2 \sin\left(\frac{\theta}{2}\right)$ Answer 2 Chloe Evans The length of the chord subtending an angle $ heta$ at the center of a unit circle can be determined using the following trigonometric approach. Consider the unit circle centered at the origin with radius 1. The chord subtends an angle $ heta$ at the center. The formula for the length of the chord is: $L = 2 sinleft(frac{ heta}{2} ight)$ Steps: Draw the radii to the endpoints of the chord, creating an isosceles triangle. The angle between the radii is $ heta$. Drop a perpendicular from the center to the chord, bisecting the angle $ heta$ and the chord. The length of the half-chord is $sinleft(frac{ heta}{2} ight)$ because of the right triangle formed. Therefore, the full length of the chord is: $L = 2 sinleft(frac{ heta}{2} ight)$ Thus, the solution is: $L = 2 sinleft(frac{ heta}{2} ight)$ Answer 3 James Taylor The length of the chord subtending an angle $ heta$ in a unit circle is given by: $L = 2 sinleft(frac{ heta}{2} ight)$ Explanation: For a unit circle, the radius $r = 1$. Chord length formula: $L = 2 sinleft(frac{ heta}{2} ight)$ Final chord length: $L = 2 sinleft(frac{ heta}{2} ight)$ [yarpp template="yarpp-template-simple" limit=6] Start Using PopAi Today R AI Video & Image R AI One-Click PPT Generation R AI PDF/DOC Reader Try for free now Powered by the world's most advanced language model, Including chatGPT-4o Follow us AI Tools AI Presentation AI ChatPDF AI Writing AI Paraphraser AI Video Generator AI Image Download Chrome iOS Android PopAi for Education Program Overview Campus Ambassador Resources Reports Quotes Papers Celebration Messages Company Term About Us Privacy Policy Contact Us © 2025 PopAi.pro 120 ROBINSON ROAD #13-01 SINGAPORE (068913) Download app and enjoy free trial Get
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https://turkarchpediatr.org/Content/files/sayilar/117/TAP_March_2022-247.pdf
Turkish Archives of Pediatrics 247 Foreign body aspiration is an important health problem frequently observed in children and adults requiring emergency intervention.1-2 Small children have a tendency toward aspira-tion when eating, crying, or playing due to weak chewing abilities.3 While it causes symptoms such as sudden respiratory distress, cough, hoarseness, and the feeling of stuck in the throat in the early period, it may result in complications such as emphysema, atelectasis, bronchi-ectasis, and pneumothorax in the later period when the diagnosis is delayed.4-5 A 10-year-old male patient, followed up with a diagnosis of cystic fibrosis (CF), was admit-ted to our pediatric emergency clinic with the complaint of cough, shortness of breath, and fever for 3 days. During the examination, there was a decrease in respiratory sounds in the lower zone of the right lung and widespread crepitant rales more pronounced in the lower zone of both lungs. His complete blood count was found to be as follows: white blood cell: 17.3 × 109/L, neutrophil: 13.1 × 109/L, and C-reactive protein: 110 mg/L (0-5). Pulmonary radiography showed an atelectasis area and infiltration in the lower zone of the right lung. Thoracic computed tomography (CT) showed consolidation areas at the level of the right lung and in the right main bronchus, a polypoid, millimetric fat density area thought to par-tially narrow the main bronchial lumen was noted. It was reported that just as this may be long to a foreign body or an intraluminal polypoid lesion. There was Pseudomonas aeruginosa in the sputum culture, which was seen to be sensitive to the treatments he received on the antibiogram. As the thoracic CT reported that the polyp-oid, millimetric fat density area thought to partially narrow the main bronchial lumen in the right main bronchus, it was decided to perform bronchoscopy, considering that the patient might have a mucus plug or foreign body in the right bronchus. Pistachios were removed from the right bronchus in bronchoscopy. He explained that he had aspirated while lying down and eating pistachio. Tracheobronchial foreign body aspiration symptoms are not specific, and patients can present with cough, wheezing, shortness of breath, fever, and pneumonia.3 Organic foreign objects cause more tissue reactions and tend to cause complications like atelectasis and air trapping.6 Due to progressive inflammation and displacement of the foreign body to the dis-tal airway, they may eventually cause complete obstruction.7 Mucus plug can cause partial or complete airway obstruction.8 The incidence of mucus plugs in children without predispos-ing factors is unknown and probably low.9 Imaging findings of a mucus plug can also mimic a foreign body.8 In a study conducted on CT findings in 27 children suspected of having for-eign bodies, the mucus plug was associated with atelectasis, atelectasis, and pneumonia or pneumonia alone.10 Since our patient was prone to form a mucus plug due to CF, and imaging findings of the mucus plug could also mimic the foreign body, it was thought that the findings were primarily due to the mucus plug, but when the pistachios were removed by bronchos-copy, it was understood that the infective process had started after pistachios aspiration besides P. aeruginosa pneumonia in our patient. Surprise Foreign Body Aspiration with Pistachios in a Patient with Cystic Fibrosis with Persistent Atelectasis on Chest Radiography Elif Arık 1, Özlem Keskin 1, Ahmet Ulusan 2, Talat Kardaş 3, Ercan Küçükosmanoğlu 1 1Department of Pediatric Allergy and Immunology, Gaziantep University, Faculty of Medicine, Gaziantep, Turkey 2Department of Thoracic Surgery, Gaziantep University, Faculty of Medicine, Gaziantep, Turkey 3Department of Pediatrics, Gaziantep University, Faculty of Medicine, Gaziantep, Turkey Content of this journal is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License. Corresponding author: Elif Arık drelif_tepe86@hotmail.com Received: October 22, 2021 Accepted: December 7, 2021 Cite this article as: Arık E, Keskin O, Ulusan A, Kardaş T, Küçükosmanoğlu E. Surprise foreign body aspiration with pistachios in a patient with cystic fibrosis with persistent atelectasis on chest radiography. Turk Arch Pediatr. 2022;57(2):247-248. SCIENTIFIC LETTER DOI: 10.5152/TurkArchPediatr.2022.21305 Pistachios Aspiration in a Patient with CF Turk Arch Pediatr 2022; 57(2): 247-248 Definitive diagnosis is made by bronchoscopy in foreign body aspirations. Anamnesis, physical examination, and radiologi-cal examinations are often sufficient to suspect foreign body aspiration. In spite of all these, consideration of foreign object aspiration in patients without any clinical findings or history but with lung problems is sufficient indication for bronchoscopy. The mucus plug that occurs in CF patients may also cause findings and radiological images that can mimic foreign body aspiration. Therefore, in the presence of persistent radiological findings, foreign body aspiration should definitely be consid-ered, even if there is a reason to explain the etiology. Peer-review: Externally peer-reviewed. Author Contributions: Concept – E.K, E.A.; Design – E.A., Ö.K.; Supervi-sion – Ö.K., E.K.; Resources – A.U., E.A.; Materials – T.K., Ö.K.; Data Col-lection and/or Processing –E.K., E.A.; Analysis and/or Interpretation – Ö.K., E.K., A.U; Literature Search – E.A., T.K.; Writing Manuscript – E.A.; Critical Review – Ö.K., E.K., E.A. Conflict of Interest: The authors have no conflict of interest to declare. Financial Disclosure: The authors declared that this study has received no financial support. REFERENCES 1. Foltran F, Ballali S, Passali FM, et al. Foreignbodies in theairways: a meta-analysis of publishedpapers. Int J Pediatr Otorhinolaryn-gol. 2012;76:S12-S19. [CrossRef] 2. Eren S, Balci AE, Dikici B, Doblan M, Eren MN. Foreign body aspira-tion in children: experience of 1160 cases. Ann Trop Paediatr. 2003;23(1):31-37. [CrossRef] 3. Sahin A, Meteroglu F, Eren S, Celik Y. Inhalation of foreign bodies in children: experience of 22 years. J Trauma Acute Care Surg. 2013;74(2):658-663. [CrossRef] 4. Smitheringale A. Management of foreign bodies of the trachea bronchial tree. In: Pearson FG, ed. Thoracic Surgery. Philadelphia: Churchil Livingstone; 1995:591-599. 5. Elhassani NB. Tracheo bronchial foreign bodies in the Middle East. A Baghdad study. J Thorac Cardiovasc Surg. 1988;96(4):621-625. [CrossRef] 6. Hegde SV, Hui PKT, Lee EY. Tracheo bronchial foreign bodies in children: imaging assessment. Semin Ultrasound CT MR. 2015;36(1):8-20. [CrossRef] 7. Tokar B, Ozkan R, Ilhan H. Tracheo bronchial foreign bodies in children: importance of accurate history and plain chest radiog-raphy in delayed presentation. Clin Radiol. 2004;59(7):609-615. [CrossRef] 8. Xue FS, Luo MP, Liao X, Liu JH, Zhang YM. Delayed endotracheal tube obstruction by mucus plug in a child. Chin Med J (Engl). 2009;122(7):870-872. 9. Debnath J, Jain NK, Adhikari KM, Shelley SK, Vaidya A, George RA. A child with respiratory distress having unilateral obstructive emphysema and contralateral opaque hemithorax on chest radi-ograph: role of multidetector computerized tomography with mul-tiplanar reconstruction and virtual bronchoscopy. Aust Radiol. 2007;51:217-220. 10. Hong WS, Im SA, Kim HL, Yoon JS. CT evaluation of airway foreign bodies in children: emphasis on the delayed diagnosis and dif-ferentiation from airway mucus plugs. Jpn J Radiol. 2013;31(1):31-38. [CrossRef] 248
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https://www.worldhistory.org/article/881/trade-in-the-phoenician-world/
Trade in the Phoenician World 34 days left Mobile App Fundraiser Quality history education gives people the tools to learn about our collective past, engage thoughtfully in civic life, recognize bias, and resist manipulation. That’s why we're launching WHE’s Mobile App — bringing reliable and unbiased history education to millions worldwide, at no cost. $4836 / $15000 ❤ Donate 51Save Article by Mark Cartwright published on Available in other languages: French, Italian, Spanish Subscribe to topic Subscribe to author Print Article PDF The Phoenicians, based on a narrow coastal strip of the Levant, put their excellent seafaring skills to good use and created a network of colonies and trade centres across the ancient Mediterranean. Their major trade routes were by sea to the Greek islands, across southern Europe, down the Atlantic coast of Africa, and up to ancient Britain. In addition, Arabia and India were reached via the Red Sea, and vast areas of Western Asia were connected to the homeland via land routes where goods were transported by caravan. By the 9th century BCE, the Phoenicians had established themselves as one of the greatest trading powers in the ancient world. Geographical Extent Trade and the search for valuable commodities necessitated the establishment of permanent trading posts and, as the Phoenician ships generally sailed close to the coast and only in daytime, regular way-stations too. These outposts became more firmly established in order to control the trade in specific commodities available at that specific site. In time, these developed further to become full colonies so that a permanent Phoenician influence eventually extended around the whole coastline of the ancient Mediterranean and the Red Sea. Their broad-bottomed single-sail cargo ships transported goods from Lebanon to the Atlantic coast of Africa, Britain, and even the Canary Islands, and brought goods back in the opposite direction, stopping at trade centres anywhere else between. Nor was trade restricted to sea routes as Phoenician caravans also operated throughout Western Asia tapping into well-established trading zones such as Mesopotamia and India. Remove Ads Advertisement Phoenician sea trade can, therefore, be divided into that for its colonies and that with fellow trading civilizations. Consequently, the Phoenicians not only imported what they needed and exported what they themselves cultivated and manufactured but they could also act as middlemen traders transporting goods such as papyrus, textiles, metals, and spices between the many civilizations with whom they had contact. They could thus make enormous gains by selling a commodity with a low value such as oil or pottery for another such as tin or silver which was not itself valued by its producers but could fetch enormous prices elsewhere. Trading Phoenicians appear in all manner of ancient sources, from Mesopotamian reliefs to the works of Homer and Herodotus, from Egyptian tomb art to the Book of Ezekiel in the Bible. The Phoenicians were the equivalent of the international haulage trucks of today, and just as ubiquitous. Follow us on YouTube! Methods of Exchange As with many other ancient civilizations the Phoenicians traded goods using a variety of methods. Prestige goods could be exchanged as reciprocal gifts but these could be more than mutual tokens of goodwill as, by bestowing on the receiver an obligation, they were a method to initiate trade partnerships. Luxury goods given as gifts may also have been a deliberate attempt by the Phoenicians to create a demand for more such items and help the Phoenicians acquire the local resources they coveted. Remove Ads Advertisement Goods were bought or sold in a relatively controlled manner where quantities and prices were fixed beforehand through the drawing up of trade agreements and treaties. Goods could be collected as a form of tribute in return for military protection or under compulsion. These were then stored in large quantities and then redistributed either locally or traded elsewhere. Goods could be bartered for and exchanged in kind on the spot. Alternatively, and perhaps the most common method employed by the Phoenicians, goods could be bought or sold in a relatively controlled manner where quantities and prices were fixed beforehand through the drawing up of trade agreements and treaties controlled by the state. The exchange value of goods was, therefore, fixed and so coinage was unnecessary, which is not to say there was no system of written arbitrary values and credit arrangements. The Phoenicians may not have produced coinage precisely because their trade was truly international and they had no use for coins which could not be used far from the place of their mint. Completely free trade where prices fluctuate due to supply and demand is a mechanism thought by some historians not to have been in operation prior to the 4th century BCE but the view is much debated amongst scholars. Phoenician trade was likely, then, carried out by state officials working on commission but also by consortiums of traders closely associated with royal households. These latter would have been high-ranking nobles, as described in Isaiah 23:8, "Tyre, the crowning city, whose merchants are princes, whose traffickers are the honourable of the earth." Perhaps from around the 8th century BCE the quantity of trade carried out by private merchants increased and the direct intervention of the state was reduced, again, the point is still subject to academic debate. The trading of goods most often took place in state-sanctioned trade centres which were generally recognised as neutral by the different regional states. The Phoenician city of Tyre is a classic example. Remove Ads Advertisement Exported Goods - Wood Phoenicia was a mere coastal strip backed by mountains. Despite the paucity of land available they did manage to produce cereals through irrigation of the arable terrain and cultivate on a limited scale such foodstuffs as olives, figs, dates, walnuts, almonds, pomegranates, plums, apricots, melons, pumpkins, cucumbers, and wine. However, the Phoenicians were most noted as exporters of wood. This commodity came from their abundant cedar and fir forests and had been traded since the beginning of recorded history. The cedar is a tall tree with a thick girth, making it ideal for timber. It also has the additional benefit of possessing an aromatic odour. Mesopotamia and Egypt were the most notable customers, the former receiving their trunks via caravan up to the Euphrates River while ships carried the wood to the African coast. The trade is recorded in reliefs of Sargon II and an inscription of Nebuchadnezzar. According to the historian George Rawlinson, Phoenician cedar wood was used by King Solomon for his celebrated temple, by Herod in Zerubbabel's Temple, and by the Ephesians for the roof of the Temple of Artemis at Ephesus, one of the Seven Wonders of the Ancient World. Textiles The other famous Phoenician export was textiles which used wool, linen yarn, cotton, and later, silk. Wool (sheep and goat) probably dominated and came from Damascus and Arabia. Linen yarn was imported from Egypt while silk came from Persia. Taking these raw materials, the Phoenicians transformed them into uniquely colourful items, especially clothes and carpets. Fine multi-coloured clothing from Phoenicia is referenced both in Homer - where Paris gives Helen a gift of the cloth prior to whisking her off to Troy - and in Egyptian art when depicting Phoenicians from Sidon. The dyed fabrics were then exported back again, for example, to Memphis where the Phoenicians even had their own quarter in the city. Cloth died purple using fluid from the murex shellfish brought the Phoenicians fame throughout the ancient world. Cloth dyed purple (actually shades ranging from pink to violet) using fluid from the Murex trunculus, Purpura lapillus, Helix ianthina, and especially the Murex brandaris shellfish brought the Phoenicians fame throughout the ancient world. Living in relatively deep water, these shell-fish were caught in baited traps suspended from floats. The dye was then extracted from thousands of putrefied shellfish left to bake in the sun. So popular were these textiles that vast deposits of the shells have been excavated on the outskirts of Sidon and Tyre and the species was all but driven to extinction along the coasts of Phoenicia. The highest quality cloth was known as Dibapha, meaning 'twice dipped' in the purple dye. The Phoenicians not only exported the dyed cloth but also the process of extracting the dye, as indicated by the shell deposits found at Phoenician colonies across the Mediterranean. Besides their vivid colours, Phoenician textiles were also famous for their fine embroidery. Popular designs included repeated motifs such as scarabs, rosettes, winged globes, lotus blossoms, and mythical monsters. Remove Ads Advertisement Glass The Phoenicians also traded glassware. The Egyptians had already been long-time producers but from the 7th century BCE the Phoenicians began to produce transparent glass, as opposed to merely opaque glassware. Important centres of glass production were Sidon, Tyre, and Sarepta. Transparent glass was used to manufacture mirrors, plates, and drinking glasses but the Phoenicians seemed to have appreciated semi-transparent coloured glass (blue, yellow, green, and brown) for their more elaborate productions as well as for jewellery and small plaques which were sewn onto clothing. Phoenician glassware, especially in the form of small perfume bottles, has been found as far afield as Cyprus, Sardinia, and Rhodes. Imported Goods The Phoenicians imported metals, especially copper from Cyprus, silver and iron from Spain, and gold from Ethiopia (and possibly Anatolia). This raw material was transformed into ornate vessels and art objects in Phoenician workshops and then exported. Tin (from Britain), lead (Scilly Isles and Spain), and brassware were also traded, the latter principally coming from Spain. Ivory was imported from either Punt or India, as was ebony, both coming to Phoenicia via Arabia. Amber came either from the Baltic or Adriatic coast and was used in Phoenician jewellery. Embroidered linen and grain were imported from Egypt and fine, worked cloth from Mesopotamia. Grain, barley, honey, and oak timber used for oars on Phoenician ships, came from Palestine. Phoenician markets also traded in slaves (from Cilicia and Phrygia but also captured by the Phoenicians themselves), sheep (Arabia), horses and mules (Armenia), goats, wool (Damascus and Arabia), coral, perfumes (Judah and Israel), agate, and precious stones such as emeralds (from Syria and Sheba). Spices came from the Arabian peninsula (some coming from distant India) and included cinnamon, calamus, cassia, ladanum, frankincense, and myrrh. Remove Ads Advertisement Legacy From the 7th century BCE the Phoenicians' trade network was eclipsed by the efforts of one of its most successful colonies - Carthage, by the Greeks, and then the Romans. But the Phoenicians had been the first Mediterranean trading superpower, and their early dominance led to those empires which followed adopting similar trading practices and even adopting Phoenician names for certain exotic goods from distant lands. The Phoenicians had dared to sail beyond the horizon and transport commodities to where they were most prized. As the prophet Isaiah (23:2) stated, "you merchants of Sidon, whose goods travelled over the sea, over wide oceans." Did you like this article? Subscribe to topic Bibliography Related Content Books Cite This Work License Editorial Review This human-authored article has been reviewed by our editorial team before publication to ensure accuracy, reliability and adherence to academic standards in accordance with our editorial policy. Remove Ads Advertisement Bibliography Aubet, M.E. The Phoenicians and the West. Cambridge University Press, 2001. Kenrick, J. Phoenicia. Forgotten Books, 2015. Moscati, S. The World of the Phoenicians Paperback. Weidenfeld & Nicolson History, 2016. Rawlinson, G. History of Phoenicia. CreateSpace Independent Publishing Platform, 2014. Book Recommendation The World of the Phoenicians (Phoenix Giants) by Moscati, Sabatino by Sabatino Moscati published by Weidenfeld & Nicolson History (1999) $63.36 World History Encyclopedia is an Amazon Associate and earns a commission on qualifying book purchases. Subscribe to this author About the Author Mark Cartwright Mark is a full-time writer, researcher, historian, and editor. Special interests include art, architecture, and discovering the ideas that all civilizations share. He holds an MA in Political Philosophy and is the WHE Publishing Director. Translations French Italian Spanish We want people all over the world to learn about history. Help us and translate this article into another language! Related Content Filters: All Definitions67 Articles22 Images6 Videos2 Collections1 Teaching Materials1 Definition Phoenicia Phoenicia was an ancient civilization composed of independent city-states... Definition Melqart Melqart (also Melkarth or Melicarthus) was an important Phoenician... Definition Tyre Tyre (in modern-day Lebanon) is one of the oldest cities in the... Definition Tyrian Purple - The Super-Expensive Dye of Antiquity Tyrian purple (aka Royal purple or Imperial purple) is a dye extracted... Definition Amber in Antiquity Amber, the fossilised resin of trees, was used throughout the ancient... Definition Canaan Canaan was the name of a large and prosperous ancient country (at... Free for the World, Supported by You World History Encyclopedia is a non-profit organization. Please support free history education for millions of learners worldwide for only $5 per month by becoming a member. Thank you! World History Encyclopedia is a non-profit organization. Please support free history education for millions of learners worldwide for only $5 per month by becoming a member. Thank you! Become a Member Donate Cite This Work APA Style Cartwright, M. (2016, April 01). Trade in the Phoenician World. World History Encyclopedia. Retrieved from Chicago Style Cartwright, Mark. "Trade in the Phoenician World." World History Encyclopedia. Last modified April 01, 2016. MLA Style Cartwright, Mark. "Trade in the Phoenician World." World History Encyclopedia. World History Encyclopedia, 01 Apr 2016, Web. 11 Aug 2025. License & Copyright Submitted by Mark Cartwright, published on 01 April 2016. The copyright holder has published this content under the following license: Creative Commons Attribution-NonCommercial-ShareAlike. This license lets others remix, tweak, and build upon this content non-commercially, as long as they credit the author and license their new creations under the identical terms. 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https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-explogfns-2009-1.pdf
Exponential and logarithm functions mc-TY-explogfns-2009-1 Exponential functions and logarithm functions are important in both theory and practice. In this unit we look at the graphs of exponential and logarithm functions, and see how they are related. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial on this topic, you should be able to: • specify for which values of a the exponential function f(x) = ax may be defined, • recognize the domain and range of an exponential function, • identify a particular point which is on the graph of every exponential function, • specify for which values of a the logarithm function f(x) = loga x may be defined, • recognize the domain and range of a logarithm function, • identify a particular point which is on the graph of every logarithm function, • understand the relationship between the exponential function f(x) = ex and the natural logarithm function f(x) = ln x. Contents 1. Exponential functions 2 2. Logarithm functions 5 3. The relationship between exponential functions and logarithm functions 9 www.mathcentre.ac.uk 1 c ⃝mathcentre 2009 1. Exponential functions Consider a function of the form f(x) = ax, where a > 0. Such a function is called an exponential function. We can take three different cases, where a = 1, 0 < a < 1 and a > 1. If a = 1 then f(x) = 1x = 1. So this just gives us the constant function f(x) = 1. What happens if a > 1? To examine this case, take a numerical example. Suppose that a = 2. f(x) = 2x f(0) = 20 = 1 f(1) = 21 = 2 f(−1) = 2−1 = 1/21 = 1 2 f(2) = 22 = 4 f(−2) = 2−2 = 1/22 = 1 4 f(3) = 23 = 8 f(−3) = 2−3 = 1/23 = 1 8 We can put these results into a table, and plot a graph of the function. x f(x) −3 1 8 −2 1 4 −1 1 2 0 1 1 2 2 4 3 8 f(x) x f(x) = 2x This example demonstrates the general shape for graphs of functions of the form f(x) = ax when a > 1. What is the effect of varying a? We can see this by looking at sketches of a few graphs of similar functions. f(x) x f(x) = 2x f(x) = 5x f(x) = 10x www.mathcentre.ac.uk 2 c ⃝mathcentre 2009 The important properties of the graphs of these types of functions are: • f(0) = 1 for all values of a. This is because a0 = 1 for any value of a. • f(x) > 0 for all values of a. This is because a > 0 implies ax > 0. What happens if 0 < a < 1? To examine this case, take another numerical example. Suppose that a = 1 2. f(x) = 1 2 x f(0) = 1 2 0 = 1 f(1) = 1 2 1 = 1 2  f(−1) = 1 2 −1 = 2 1 1 = 2 f(2) = 1 2 2 = 1 4  f(−2) = 1 2 −2 = 2 1 2 = 4 f(3) = 1 2 3 = 1 8  f(−3) = 1 2 −3 = 2 1 3 = 8 We can put these results into a table, and plot a graph of the function. x f(x) −3 8 −2 4 −1 2 0 1 1 1 2 2 1 4 3 1 8 f(x) x f(x) = ( ) x 1 2 This example demonstrates the general shape for graphs of functions of the form f(x) = ax when 0 < a < 1. What is the effect of varying a? Again we can see by looking at sketches of a few graphs of similar functions. f(x) x f(x) = ( ) x 1 2 f(x) = ( ) x 1 10 f(x) = ( ) x 1 5 www.mathcentre.ac.uk 3 c ⃝mathcentre 2009 The important properties of the graphs of these types of functions are: • f(0) = 1 for all values of a. This is because a0 = 1 for any value of a. • f(x) > 0 for all values of a. This is because a > 0 implies ax > 0. Notice that these properties are the same as when a > 1. One interesting thing that you might have spotted is that f(x) = ( 1 2)x = 2−x is a reflection of f(x) = 2x in the f(x) axis, and that f(x) = ( 1 5)x = 5−x is a reflection of f(x) = 5x in the f(x) axis. f(x) x f(x) = ( ) x 1 2 f(x) = ( ) x 1 5 f(x) = 2x f(x) = 5x In general, f(x) = (1/a)x = a−x is a reflection of f(x) = ax in the f(x) axis. A particularly important example of an exponential function arises when a = e. You might recall that the number e is approximately equal to 2.718. The function f(x) = ex is often called ‘the’ exponential function. Since e > 1 and 1/e < 1, we can sketch the graphs of the exponential functions f(x) = ex and f(x) = e−x = (1/e)x. f(x) x f(x) = ex f(x) = e−x www.mathcentre.ac.uk 4 c ⃝mathcentre 2009 Key Point A function of the form f(x) = ax (where a > 0) is called an exponential function. The function f(x) = 1x is just the constant function f(x) = 1. The function f(x) = ax for a > 1 has a graph which is close to the x-axis for negative x and increases rapidly for positive x. The function f(x) = ax for 0 < a < 1 has a graph which is close to the x-axis for positive x and increases rapidly for decreasing negative x. For any value of a, the graph always passes through the point (0, 1). The graph of f(x) = (1/a)x = a−x is a reflection, in the vertical axis, of the graph of f(x) = ax. A particularly important exponental function is f(x) = ex, where e = 2.718 . . .. This is often called ‘the’ exponential function. 2. Logarithm functions We shall now look at logarithm functions. These are functions of the form f(x) = loga x where a > 0. We do not consider the case a = 1, as this will not give us a valid function. What happens if a > 1? To examine this case, take a numerical example. Suppose that a = 2. Then f(x) = log2 x means 2f(x) = x . An important point to note here is that, regardless of the argument, 2f(x) > 0. So we shall consider only positive arguments. f(1) = log2 1 means 2f(1) = 1 so f(1) = 0 f(2) = log2 2 means 2f(2) = 2 so f(2) = 1 f(4) = log2 4 means 2f(4) = 4 so f(4) = 2 f( 1 2) = log2( 1 2) means 2f(1 2 ) = 1 2 = 2−1 so f( 1 2) = −1 f( 1 4) = log2( 1 4) means 2f(1 4 ) = 1 4 = 2−2 so f( 1 4) = −2 We can put these results into a table, and plot a graph of the function. www.mathcentre.ac.uk 5 c ⃝mathcentre 2009 x f(x) 1 4 −2 1 2 −1 1 0 2 1 4 2 f(x) x f(x) = log2 x This example demonstrates the general shape for graphs of functions of the form f(x) = loga x when a > 1. What is the effect of varying a? We can see by looking at sketches of a few graphs of similar functions. For the special case where a = e, we often write ln x instead of loge x. f(x) x f(x) = log2 x f(x) = log5 x f(x) = loge x = ln x The important properties of the graphs of these types of functions are: • f(1) = 0 for all values of a; • we must have x > 0 for all values of a. What happens if 0 < a < 1? To examine this case, take another numerical example. Suppose that a = 1 2. Then f(x) = log1/2 x means 1 2 f(x) = x . An important point to note here is that, regardless of the argument, 1 2 f(x) > 0. So we shall consider only positive arguments. www.mathcentre.ac.uk 6 c ⃝mathcentre 2009 f(x) = 1 2 x f(1) = log1/2 1 means 1 2 f(1) = 1 so f(1) = 0 f(2) = log1/2 2 means 1 2 f(2) = 2 = 1 2 −1 so f(2) = −1 f(4) = log1/2 4 means 1 2 f(4) = 4 = 1 2 −2 so f(4) = −2 f( 1 2) = log1/2( 1 2) means 1 2 f(1 2 ) = 1 2 so f( 1 2) = 1 f( 1 4) = log1/2( 1 4) means 1 2 f(1 4 ) = 1 4 = 1 2 2 so f( 1 4) = 2 We can put these results into a table, and plot a graph of the function. x f(x) 1 4 2 1 2 1 1 0 2 −1 4 −2 f(x) x f(x) = log1/2 x This example demonstrates the general shape for graphs of functions of the form f(x) = loga x when 0 < a < 1. What is the effect of varying a? Again we can see by looking at sketches of a few graphs of similar functions. f(x) x f(x) = log1/2 x f(x) = log1/5 x f(x) = log1/e x www.mathcentre.ac.uk 7 c ⃝mathcentre 2009 The important properties of the graphs of these types of functions are: • f(1) = 0 for all values of a; • we must have x > 0 for all values of a. An interesting thing that you might well have spotted is that f(x) = log1/5 x is a reflection of f(x) = log5 x in the x-axis and f(x) = log1/2 x is a reflection of f(x) = log2 x in the x-axis. f(x) x f(x) = log1/2 x f(x) = log1/5 x f(x) = log2 x f(x) = log5 x Generally, f(x) = log1/a x is a reflection of f(x) = loga x in the x-axis. Key Point A function of the form f(x) = loga x (where a > 0 and a ̸= 1) is called a logarithm function. The function f(x) = loga x for a > 1 has a graph which is close to the negative f(x)-axis for x < 1 and increases slowly for positive x. The function f(x) = loga x for 0 < a < 1 has a graph which is close to the positive f(x)-axis for x < 1 and decreases slowly for positive x. For any value of a, the graph always passes through the point (1, 0). The graph of f(x) = log1/a x is a reflection, in the horizontal axis, of the graph of f(x) = loga x. A particularly important logarithm function is f(x) = loge x, where e = 2.718 . . .. This is often called the natural logarithm function, and written f(x) = ln x. www.mathcentre.ac.uk 8 c ⃝mathcentre 2009 3. The relationship between exponential functions and log-arithm functions We can see the relationship between the exponential function f(x) = ex and the logarithm function f(x) = ln x by looking at their graphs. f(x) x f(x) = ln x f(x) = ex f(x) = x You can see straight away that the logarithm function is a reflection of the exponential function in the line represented by f(x) = x. In other words, the axes have been swapped: x becomes f(x), and f(x) becomes x. Key Point The exponential function f(x) = ex is the inverse of the logarithm function f(x) = ln x. Exercises 1. Sketch the graph of the function f(x) = ax for the following values of a, on the same axes. (a) a = 3 (b) a = 6 (c) a = 1 (d) a = 1 3 (e) a = 1 6 2. Sketch the graph of the function f(x) = loga x for the following values of a, on the same axes. (a) a = 3 (b) a = 6 (c) a = 1 3 (d) a = 1 6 www.mathcentre.ac.uk 9 c ⃝mathcentre 2009 3. For each of the following pairs of functions, state whether the graphs are related by a reflection in the x-axis, a reflection in the f(x)-axis, a reflection in the line f(x) = x, a reflection in the line f(x) = −x, or that the graphs are not related by any of these reflections. (a) f(x) = 3x and f(x) = 1 3 x (b) f(x) = log6 x and f(x) = 6x (c) f(x) = log6 x and f(x) = 1 6 x (d) f(x) = log1/3 x and f(x) = log3 x (e) f(x) = 1 3 x and f(x) = 1 6 x Answers 1. f(x) x f(x) = ( ) x 1 3 f(x) = ( ) x 1 6 f(x) = 3x f(x) = 6x f(x) = 1x 2. f(x) x f(x) = log1/3 x f(x) = log1/6 x f(x) = log3 x f(x) = log6 x www.mathcentre.ac.uk 10 c ⃝mathcentre 2009 3. (a) Reflect in the f(x)-axis (b) Reflect in the line f(x) = x (c) Not related by any of these reflections (d) Reflect in the x-axis (e) Not related by any of these reflections www.mathcentre.ac.uk 11 c ⃝mathcentre 2009
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http://labman.phys.utk.edu/phys221core/modules/m5/conservation_of_momentum.html
Conservation of momentum Consider two interacting objects. If object 1 pushes on object 2 with a force F = 10 N for 2 s to the right, then the momentum of object 2 changes by 20 Ns = 20 kg m/s to the right. By Newton's third law object 2 pushes on object 1 with a force F = 10 N for 2 s to the left. The momentum of object 1 changes by 20 Ns = 20 kg m/s to the left. The total momentum of both objects does not change. For this reason we say that the total momentum of the interacting objects is conserved. Newton's third law implies that the total momentum of a system of interacting objects not acted on by outside forces is conserved. The total momentum in the universe is conserved. The momentum of a single object, however, changes when a net force acts on the object for a finite time interval. Conversely, if no net force acts on an object, its momentum is constant. For a system of objects, a component of the momentum along a chosen direction is constant, if no net outside force with a component in this chosen direction acts on the system. Collisions In collisions between two isolated objects Newton's third law implies that momentum is always conserved. In collisions, it is assumed that the colliding objects interact for such a short time, that the impulse due to external forces is negligible. Thus the total momentum of the system just before the collision is the same as the total momentum just after the collision. Collisions in which the kinetic energy is also conserved, i.e. in which the kinetic energy just after the collision equals the kinetic energy just before the collision, are called elastic collisions. In these collisions no ordered energy is converted into thermal energy. Collisions in which the kinetic energy is not conserved, i.e. in which some ordered energy is converted into internal energy, are called inelastic collisions. If the two objects stick together after the collision and move with a common velocity vf, then the collision is said to be perfectly inelastic. Note:In collisions between two isolated objects momentum is always conserved.Kinetic energy is only conserved in elastic collisions. We always have m1v1i + m2v2i = m1v1f + m2v2f.Only for elastic collisions do we also have ½m1v1i2 + ½m2v2i2 = ½m1v1f2 + ½m2v2f2. Problem: If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision? Is it possible for one to be at rest after the collision? Explain! Solution: Problem: A 10 g bullet is stopped in a block of wood (m = 5 kg). The speed of the bullet-wood combination immediately after the collision is 0.6 m/s. What was the original speed of the bullet? Solution: Problem: Two cars of equal mass and equal speeds collide head on. Do they experience a greater force if the collision is elastic or perfectly inelastic and they stick together? Solution: Problem: A 90 kg fullback running east with a speed of 5 m/s is tackled by a 95 kg opponent running north with a speed of 3 m/s. If the collision is perfectly inelastic, calculate the speed and the direction of the players just after the tackle. Solution: Problem: A 30,000 kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal into it. What is the final velocity of the loaded freight car? Solution: Problem: After a completely inelastic collision between two objects of equal mass, each having initial speed v, the two move off together with speed v/3. What was the angle between their initial directions? Solution: Problem: The mass of the blue puck is 20% greater than the mass of the green one. Before colliding, the pucks approach each other with equal and opposite momenta, and the green puck has an initial speed of 10 m/s. Find the speed of the pucks after the collision, if half the kinetic energy is lost during the collision. Solution: Embedded Question 1 During a visit to the International Space Station, an astronaut was positioned motionless in the center of the station, out of reach of any solid object on which he could exert a force. Suggest a method by which he could move himself away from this position, and explain the physics involved. Discuss this with your fellow students in the discussion forum! Embedded Question 2 Football coaches advise players to block, hit, and tackle with their feet on the ground rather than by leaping through the air. Using the concepts of momentum, work, and energy, explain how a football player can be more effective with his feet on the ground. Discuss this with your fellow students in the discussion forum! External links:
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https://appspenang.uitm.edu.my/sigcs/2025-1/Articles/2025V1_PAPER12.pdf
e-ISBN : 978-629-98755-5-0 SIG : e-Learning@CS Publication Date : 24 – Mac - 2025 76 CLUSTER SAMPLING IN EDUCATIONAL RESEARCH: A PRACTICAL APPROACH Nurhafizah Ahmad1, Fadzilawani Astifar Alias2 and Siti Asmah Mohamed3 nurha9129@uitm.edu.my1, fadzilawani.astifar@uitm.edu.my2, sitiasmah109@uitm.edu.my3 1,2,3Jabatan Sains Komputer & Matematik (JSKM), Universiti Teknologi MARA Cawangan Pulau Pinang, Malaysia Corresponding author ABSTRACT Cluster sampling is a widely employed probability sampling technique in educational research, particularly useful for large-scale studies where logistical and financial constraints limit the feasibility of simple random sampling. This method involves selecting entire clusters, such as schools, classrooms, or districts, rather than individual participants, making it ideal for educational settings with naturally occurring group structures. By streamlining data collection processes, cluster sampling enhances efficiency while ensuring representative sampling within a defined population. This paper explores the concept, significance, and practical application of cluster sampling in educational research. It discusses its advantages and limitations and provides an extensive review of empirical studies that have successfully applied this technique. Furthermore, it outlines the procedural steps required for effective implementation, ensuring methodological rigor and minimizing bias. The discussion highlights how cluster sampling has facilitated large-scale educational assessments, policy evaluations, and pedagogical research, reinforcing its value as a methodological tool in contemporary educational research. Keywords: cluster sampling, educational research, sampling method, statistical analysis Introduction Educational research often requires the collection and analysis of data from large and diverse populations. However, obtaining data from every individual in a target population can be prohibitively expensive and time-consuming, particularly when research spans multiple institutions, regions, or even countries. Traditional sampling methods, such as simple random sampling, can become impractical when dealing with educational settings where individuals naturally exist within groups, such as schools, classrooms, or districts. In such contexts, cluster sampling provides an efficient and cost-effective alternative by selecting entire groups, or clusters, for study instead of sampling individuals independently. This approach reduces logistical constraints while maintaining the representativeness of the sample. By leveraging naturally occurring clusters, researchers can efficiently collect data across large populations without compromising statistical integrity. This paper delves into the theoretical underpinnings of cluster sampling, its applications in educational research, and the methodological steps required for its effective implementation. e-ISBN : 978-629-98755-5-0 SIG : e-Learning@CS Publication Date : 24 – Mac - 2025 77 Understanding Cluster Sampling Cluster sampling is a probability sampling technique where a population is divided into distinct clusters, and entire clusters are randomly selected for inclusion in a study. Each cluster should ideally represent the broader population to ensure generalizability. Unlike stratified sampling, which involves selecting individual elements from each subgroup, cluster sampling simplifies the selection process by treating whole groups as units of analysis. There are two main types of cluster sampling: single-stage and multi-stage sampling. In single-stage cluster sampling, all individuals within selected clusters are included in the study. In contrast, multi-stage cluster sampling involves additional randomization within selected clusters, further refining the sample to enhance representativeness. The choice between these approaches depends on study objectives, resource availability, and population heterogeneity. Despite its efficiency, cluster sampling introduces potential challenges such as intra-cluster correlation, where similarities among individuals within a cluster can reduce statistical precision. However, with appropriate design considerations, these limitations can be mitigated, making cluster sampling a robust tool for educational research (Lohr, 2021). Application of Cluster Sampling in Educational Research Cluster sampling has been extensively applied in educational research, particularly in large-scale assessments, policy evaluations, and pedagogical studies. By using naturally occurring groups, such as schools and classrooms, researchers can conduct large-scale studies more efficiently while maintaining statistical validity. Below, we explore how cluster sampling has been applied in key research studies, demonstrating its practical utility and methodological rigor in educational research. One of the most prominent examples of cluster sampling in educational research is the Programme for International Student Assessment (PISA), an international large-scale educational survey conducted by the Organisation for Economic Co-operation and Development (OECD, 2019). PISA assesses the knowledge and skills of 15-year-old students across various countries in subjects such as reading, mathematics, and science. PISA employs a two-stage cluster sampling approach. In the first stage, a sample of schools is randomly selected from each participating country, ensuring a diverse and representative selection of institutions that reflect regional and socioeconomic variations. In the second stage, a random sample of students within each selected school is drawn. This method allows researchers to make cross-national comparisons without the logistical burden of individually sampling students from an entire national population. The use of cluster sampling in PISA ensures cost efficiency and operational feasibility in global educational assessments. However, it also requires careful statistical adjustments, such as weighting techniques, to account for intra-cluster correlation, ensuring that results accurately reflect the national e-ISBN : 978-629-98755-5-0 SIG : e-Learning@CS Publication Date : 24 – Mac - 2025 78 and international student populations. The findings from PISA have informed educational policies worldwide, shaping curriculum development, teaching strategies, and funding allocations in numerous countries. The National Assessment of Educational Progress (NAEP), often referred to as the "Nation’s Report Card," is another large-scale educational study that relies on cluster sampling. NAEP assesses the academic proficiency of students in the United States in subjects such as mathematics, reading, and science (NCES, 2021). Similar to PISA, NAEP uses a multi-stage cluster sampling design. In the first stage, schools are selected as primary clusters based on stratification criteria such as geographical location, school size, and student demographics. In the second stage, students within selected schools are randomly chosen to participate in the assessment. The cluster sampling approach allows NAEP to maintain a nationally representative sample without testing every student in the country. This method reduces data collection costs while providing accurate estimates of student achievement trends over time. NAEP results are used by policymakers, educators, and researchers to evaluate the effectiveness of educational reforms, track achievement gaps, and guide policy decisions at the federal and state levels. The Trends in International Mathematics and Science Study (TIMSS) is another major international educational assessment that relies on cluster sampling. TIMSS measures the mathematics and science proficiency of fourth- and eighth-grade students across multiple countries (Mullis et al., 2019). TIMSS employs a two-stage cluster sampling process similar to PISA. In the first stage, schools are randomly selected within each country, ensuring diversity in terms of location, funding levels, and student backgrounds. In the second stage, entire classrooms within the selected schools are chosen, rather than individual students. This means that all students within a selected classroom participate in the assessment. By clustering students within classrooms, TIMSS reduces logistical challenges and standardizes testing conditions, making administration more efficient. The results from TIMSS are widely used by governments and international organizations to compare educational performance globally, develop curriculum improvements, and inform teacher training programs. A study by Gustafsson (2007) investigated school effectiveness in Sweden using cluster sampling. The research aimed to explore the impact of school-level characteristics on student achievement and educational outcomes. In this study, schools were chosen as the primary clusters, ensuring that data collection captured variations in teaching practices, administrative support, and student demographics. By selecting entire schools rather than individual students, the study was able to examine how institutional factors, such as teacher-student ratios and school funding, influenced academic performance. The findings provided valuable insights into the role of school environments in shaping student success. The study also demonstrated how cluster sampling can be used to assess e-ISBN : 978-629-98755-5-0 SIG : e-Learning@CS Publication Date : 24 – Mac - 2025 79 school-level differences in educational effectiveness, guiding policymakers in resource allocation and educational planning. A study by Dunn et al. (2019) examined how class size affects student performance in Canadian primary schools. This study utilized classrooms as the cluster units, allowing researchers to compare different teaching conditions across multiple schools. By randomly selecting classrooms within various schools, the study ensured that differences in instructional methods, student-teacher interactions, and peer effects were systematically captured. The cluster sampling approach was particularly beneficial because it accounted for the fact that students in the same classroom share common educational experiences. Findings from this study contributed to ongoing debates on optimal class sizes, informing school district policies regarding teacher assignments, classroom resources, and curriculum adaptations. Cluster sampling has also been used to evaluate the effectiveness of teacher training programs. A study by Goldhaber et al. (2020) examined the impact of professional development programs on teaching effectiveness by selecting entire school districts as clusters. In this research, rather than sampling individual teachers, entire districts were chosen, ensuring that variations in district-level training policies, resource availability, and administrative support were captured. The study compared districts that implemented intensive professional development programs with those that did not, analyzing the effects on student learning outcomes. The results provided crucial evidence on the importance of continuous professional development for educators, influencing teacher training policies at both state and national levels. A study by Fryer (2014) evaluated the impact of merit pay systems on teacher performance and student achievement in the United States. The research used cluster sampling by selecting entire schools that implemented merit pay policies and comparing them to randomly selected control schools. By analyzing full schools rather than individual teachers, the study ensured that findings reflected the broader institutional impact of merit-based compensation, rather than just individual teacher responses. The study found mixed results, with some schools benefiting from merit pay while others showed no significant changes in student outcomes. The findings contributed to the broader discussion on performance-based incentives in education, shaping future policy decisions. Longitudinal studies in education also benefit from cluster sampling. The Early Childhood Longitudinal Study (ECLS), conducted by the U.S. Department of Education, follows cohorts of students from early childhood through secondary school to track their academic and social development (NCES, 2021). ECLS uses schools as clusters, allowing researchers to follow students within structured learning environments while minimizing attrition rates. The use of cluster sampling in ECLS enables researchers to analyze long-term trends in student development, helping shape early childhood education policies and interventions. e-ISBN : 978-629-98755-5-0 SIG : e-Learning@CS Publication Date : 24 – Mac - 2025 80 Process of Sampling Using Cluster Sampling The implementation of cluster sampling in educational research follows a structured process to ensure methodological rigor and representativeness. The first step involves defining the target population and identifying natural clusters within it. In educational settings, these clusters often include schools, classrooms, or districts. Once the clusters are established, researchers develop a sampling frame and determine the sampling approach—whether single-stage or multi-stage cluster sampling. In single-stage cluster sampling, researchers randomly select a subset of clusters and include all individuals within them. This approach is particularly useful when intra-cluster variability is high, ensuring a diverse representation of the population. In multi-stage cluster sampling, an additional layer of randomization occurs within selected clusters, further refining the sample. For instance, after selecting schools as primary clusters, researchers may randomly select specific classrooms or grade levels within those schools to participate in the study. Following the selection of clusters, researchers conduct data collection while ensuring adherence to ethical considerations, such as obtaining informed consent from participants and maintaining data confidentiality. Statistical adjustments, such as weighting techniques, are often applied during data analysis to account for clustering effects and enhance the accuracy of population-level inferences. Proper methodological execution ensures that findings remain valid and generalizable despite the inherent clustering of the sample. Conclusion Cluster sampling remains a fundamental methodological tool in educational research, enabling the efficient collection of large-scale, representative data. Its application in studies such as PISA, NAEP, TIMSS, and longitudinal research highlights its effectiveness in large-scale assessments. Moreover, school effectiveness research, class size studies, teacher training evaluations, and policy analysis further demonstrate its versatility in addressing key educational questions. While challenges such as intra-cluster correlation exist, careful study design and statistical adjustments help mitigate these issues. As educational research continues to evolve, cluster sampling remains an essential approach for conducting rigorous, policy-relevant, and impactful studies that contribute to the advancement of education worldwide. e-ISBN : 978-629-98755-5-0 SIG : e-Learning@CS Publication Date : 24 – Mac - 2025 81 References: Creswell, J. W., & Creswell, J. D. (2017). Research design: Qualitative, quantitative, and mixed methods approaches (5th ed.). Sage Publications. Goldstein, H. (2011). Multilevel statistical models (4th ed.). Wiley. Retrieved from Graham, S., Harris, K. R., & Mason, L. H. (2014). Improving writing in the disciplines: A research synthesis. Educational Psychology Review, 26(1), 1-21. Hox, J. J. (2010). Multilevel analysis: Techniques and applications (2nd ed.). Routledge. Retrieved from Keele, L., & Zubizarreta, J. R. (2017). Assessing the impact of new curricular approaches through cluster sampling. Educational Research Quarterly, 40(2), 13-26. Leithwood, K., Seashore Louis, K., Anderson, S., & Wahlstrom, K. (2004). How leadership influences student learning. Center for Applied Research and Educational Improvement, University of Minnesota. Retrieved from Levy, P. S., & Lemeshow, S. (2013). Sampling of populations: Methods and applications (4th ed.). Wiley. Lohr, S. L. (2019). Sampling: Design and analysis (2nd ed.). Chapman and Hall/CRC. OECD. (2018). The survey of adult skills: Reader's companion (2nd ed.). OECD Publishing. Selwyn, N. (2011). Education and technology: Key issues and debates. Continuum. Retrieved from Snijders, T. A. B., & Bosker, R. J. (2012). Multilevel analysis: An introduction to basic and advanced multilevel modeling (2nd ed.). Sage Publications. Thompson, S. K. (2012). Sampling (3rd ed.). Wiley. Retrieved from
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https://tutorial.math.lamar.edu/problems/alg/integerexponents.aspx
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https://utp.ac.pa/documentos/2014/pdf/10_Energi_a_magne_tica_1.pdf
29 Prisma T ecnológico | Vol. 4, n.° 1, edición 2013. Tecnología a fondo Almacenamiento de energía magnética por superconducción Guadalupe G. González Universidad Tecnológica de Panamá guadalupe.gonzalez@utp.ac.pa Resumen: en este artículo presentamos una de las tecnologías de almacenamiento de energía actualmente utilizadas en sistemas de potencia, Almacenamiento de Energía por Superconducción. Primero, presentamos una pequeña reseña histórica sobre la superconductividad, sus orígenes y primeras aplicaciones, ya que es la base de este sistema de almacenamiento. Luego, explicamos los principios básicos de su funcionamiento y detallamos sus componentes principales. Finalmente, presentamos su aplicación en sistemas de potencia. Palabras claves: almacenamiento de energía, electromagnetismo, sistema de potencia, SMES, superconductor. Title: Superconducting Magnetic Energy Storage. Abstract: in this article we present one of the energy storage technologies currently used in power systems, Energy Storage by Superconduction. First, we present a small historical review on superconductivity, its origins and first applications, since it is the base of this storage system. Then, we explain the basic principles of operation and detail its main components. Finally, we present its application in power systems. Keywords: electromagnetism, energy storage, power system, SMES, superconductor. 1. Introducción E l mundo industrializado gira alrededor de energía. En los últimos años, casi el 90% de la energía primaria en el mundo proviene de derivados del petróleo, ya sea en forma de carbón, petróleo crudo o gas natural . Dado que éstos no son renovables no podemos depender de ellos como fuentes primarias de energía. En las últimas décadas se han incrementado los esfuerzos para reducir el uso del petróleo a nivel mundial. Hemos visto cómo se han desarrollado los sistemas de energías renovables para generación eléctrica así como la comercialización de vehículos eléctricos, híbridos y de combustible flexible, entre otros; sin embargo, todas estas tecnologías parecen tener un punto débil en común: el sistema de almacenamiento de energía. En los sistemas de potencia con generación eólica o solar, por ejemplo, la energía se extrae de manera intermitente; ya sea debido a variaciones en la velocidad del viento o a la presencia de nubes durante el día. Estas fluctuaciones pueden afectar significativamente el sistema de potencia por consiguiente, se necesitan sistemas de almacenamiento capaces de almacenar grandes cantidades de energía que puedan amortiguar dichas fluctuaciones. También, se necesita asegurar la extracción de la energía en cualquier momento que sea posible no sólo cuando la carga lo requiera. Finalmente, se requieren métodos de almacenamiento capaces de satisfacer variaciones en la demanda [2,3]. Para satisfacer las necesidades previamente mencionadas, se han desarrollado diferentes tipos de sistemas de almacenamiento, entre ellos: térmico, químico, electroquímico, de aire comprimido, mecánico con volantes de inercia, bombeo hidroeléctrico, banco de capacitores y por superconducción, siendo esta última el tema de este artículo. Los sistemas de almacenamiento de energía magnética por superconducción (SMES – Superconducting Magnetic Energy Storage) fueron diseñados originalmente para satisfacer variaciones en la demanda diurna. Quizás su mejor característica es que es altamente eficiente; se ha estimado que una unidad puede tener un 90% de eficiencia mientras que los sistemas de almacenamiento por bombeo hidroeléctrico, las baterías y las volantes de inercia tienen una eficiencia típica del 60-70% . 2. Reseña histórica La superconductividad fue descubierta en 1911 por Heike Kamerlingh Onnes, quien se encontraba estudiando la resistencia del mercurio sólido a temperaturas criogénicas, utilizando helio líquido como refrigerante . Alrededor de los años 60’s, científicos americanos, japoneses y europeos realizaron los primeros pasos en la creación de almacenadores de energía magnética por superconducción. Pero no fue hasta 1971, en el Centro de Superconductividad Aplicada de la Universidad de Wisconsin, cuando Peterson y Boom inventaron el sistema de SMES tal y como lo conocemos hoy en día. Después de esto, distintas compañías y centros de investigación han desarrollado y diseñado SMES para su uso en redes eléctricas. 3. Almacenamiento de energía magnética por superconducción Las unidades de almacenamiento de energía magnética por superconducción (SMES) almacenan energía de la misma forma que lo haría un inductor convencional. Ambos, almacenan energía en el campo magnético creado por las corrientes que fluyen a través de un alambre bobinado. La principal diferencia radica en que en el SMES, una corriente directa fluye a través de un alambre superconductor; esto significa que el alambre se encuentra a temperaturas criogénicas y no muestra resistencia conductiva alguna. El hecho que no exista resistencia óhmica en el alambre implica que no hay disipación térmica, por consiguiente, la energía puede almacenarse en el SMES virtualmente por tiempo indefinido hasta que sea requerida. Dado que la energía es almacenada como corriente circulatoria, puede extraerse de las unidades SMES con una respuesta casi instantánea siendo entregada o almacenada en 30 Prisma T ecnológico | Vol. 4, n.° 1, edición 2013. El contenido energético en un campo electromagnético es determinado por la corriente que fluye a través de las espiras de una bobina magnética y puede ser calculado con (1). (1) periodos que varían de fracciones de segundos a algunas horas . Una unidad típica de almacenamiento de energía por superconducción consta principalmente de: la bobina superconductora, el sistema de refrigeración y la interfaz eléctrica. 3.1. Bobina superconductora Los superconductores son capaces de transportar altos niveles de corrientes en la presencia de altos niveles de campos magnéticos a bajas temperaturas con cero resistencia al flujo de corriente eléctrica, a menos que sus valores críticos: temperatura (Tc), densidad de flujo magnético (Bc) y densidad de corriente (Ic), sean excedidos. Los materiales que exhiben superconductividad han ido creciendo en número y variedad, pero la cantidad de superconductores utilizados en aplicaciones prácticas y comerciales todavía es limitada, siendo la aleación de Niobio-Titanio (NbTi) la más utilizada en aplicaciones a altos niveles de potencia . En la Tabla 1 podemos ver una lista de superconductores con sus valores críticos de temperatura, densidad magnética y su densidad energética (Wm) . Tabla 1. Lista de Superconductores Superconductor Tc [K] Bc [T] Wm [J/m3] Metales Niobio (Nb) 9.26 0.82 2.68E+05 Tántalo (Ta) 4.48 0.30 3.58E+04 Vanadio (V) 5.03 1 3.98E+05 No-Metales C6Ca 11.5 0.95 3.59E+05 Diamante:B 11.4 4 6.37E+06 In2O3 3.30 3 3.58E+06 Si:B 0.40 0.40 6.37E+04 Aleaciones Binarias MgB2 39 74 2.18E+09 Nb3Ge 23.2 37 5.45E+08 Nb3Sn 18.3 30 3.58E+08 NbTi 10 15 8.95E+07 donde Wm es la energía almacenada [Joules]. L es la inductancia [Henrios]. i es la corriente eléctrica [Amperios]. Wm = 1 2 Li2 1. Tubería de Helio 2. Espira superconductora 3. Anillo de refuerzo 4. Cámara de vacío 5. Columna de soporte Figura 1. Pictórico de una bobina superconductora dentro de un contenedor criogénico de Helio, con capacidad de almacenar 100 MJ de energía. Figura modificada de la versión original , traducida al español. Tecnología a fondo Dado que la densidad de campo magnético en materiales ferromagnéticos no sobrepasa los 3 Teslas, las bobinas utilizadas para almacenar energía magnética por superconducción son usualmente colocadas en aire o al vacío con permeabilidad μ=μo=4πx10-7 [H/m]. Para obtener altos valores de energía (Wm) con una corriente (i) limitada por el superconductor utilizado es necesario incrementar la inductancia; lo cual se puede hacer utilizando la geometría adecuada. Existen tres configuraciones en el diseño de SMES: • Solenoide sencillo con forma circular. • Conexión en serie de solenoides coaxiales. • Toroide de forma circular, ovalado o D comprendido por una serie de espiras conectadas en series. Cabe señalar que la configuración del inductor afecta la masa total de la estructura del SMES. Inductores con configuración toroidal compuesto por espiras sencillas tienen un campo magnético externo mínimo, lo cual es ideal para no afectar a los sistemas de navegación, la salud de las personas y las líneas de transmisión, pero utilizan aproximadamente el doble de superconductor, mientras que la configuración de solenoide sencillo es superior en términos de energía almacenada por peso. 3.2. Sistema de enfriamiento El sistema de enfriamiento de un SMES está compuesto por el refrigerador, en donde se prepara el refrigerante y el contendedor criogénico en donde reposa la bobina superconductora para ser refrigerada y aislada térmicamente del medio ambiente (ver Figura.1). El sistema de enfriamiento utiliza normalmente Helio como refrigerante ya sea como baño de Helio o por circulación forzada. Éste remueve todo el calor que entra al contenedor criogénico y por consiguiente asegura que la temperatura del superconductor no exceda la temperatura crítica. Dado que la planta de refrigeración posee una eficiencia límite, el calor que penetra por las tuberías, los soportes mecánicos y por radiación deben ser tan bajo como sea posible; para asegurar esto, se provee de un enfriamiento intermedio llamado “escudo térmico”. La refrigeración de la bobina y el aislamiento térmico son problemas técnicos de extrema dificultad dado a las bajas temperaturas (alrededor de 1.8 K) que se necesitan para mantener el superconductor trabajando de manera eficiente. El refrigerador consume energía eléctrica y por consiguiente disminuye la eficiencia del SMES. Un sistema de refrigeración típico requiere aproximadamente 1.5 kW por mega watt-hora de energía almacenada . 31 Prisma T ecnológico | Vol. 4, n.° 1, edición 2013. 3.3. Interfaz eléctrica La interfaz eléctrica entre el inductor superconductor y el sistema de potencia es un convertidor. El mismo es un rectificador/inversor que cambia la corriente alterna proveniente de la red a la corriente directa que fluye continua en las bobinas. Para cargar o descargar el inductor, el voltaje, a través de las bobinas, se hace positivo o negativo. Cuando la unidad está en reposo, independientemente del nivel de energía almacenado, la corriente se mantiene constante y el voltaje promedio, a través de las bobinas superconductoras, es cero . La configuración típica de un convertidor para esta aplicación comprende dos puentes de tiristores de 6 pulsos, conectados en serie a la bobina superconductora en la parte directa del puente y acoplados, en la parte alterna, al sistema de potencia a través de un transformador. Las pérdidas correspondientes al convertidor de estado sólido se estiman alrededor del 3 al 8% del total de la energía almacenada . 4. Almacenadores de energía magnética por superconducción en sistemas de potencia En la Tabla 2, podemos ver algunos parámetros típicos de un sistema de almacenamiento por superconducción . Los sistemas SMES son capaces de almacenar de 1 MW a 10 MW. La Figura 2 muestra una comparación entre la capacidad de almacenamiento y el tiempo de descarga de distintos sistemas de almacenamiento de energía. Tecnología a fondo Tabla 2. Parámetros típicos de un SMES Total de energía almacenada 10000-13000 MWh Energía disponible 9000-10000 MWh Tiempo de descarga 5-12 h Potencia máxima 1000-2500 MW Corriente máxima 50-300 kA Densidad de campo máximo 4-6 T Diámetro medio de la bobina 300 m Altura total de la bobina 80-100 m Profundidad media debajo de la superficie 300-400 m Eficiencia 85-90 % Pérdidas en el convertidor 2% de la potencia Potencia del refrigerador 20-30 MW Asumiendo un ciclo completo de carga/descarga al día. Dado que los sistemas de almacenamiento de energía magnética por superconducción son altamente eficientes y responden rápidamente a las variaciones de la demanda, pueden ser de gran utilidad a los sistemas de potencia ya que: • tienen la capacidad de proveer energía al sistema (spinning reserve) si se presenta una pérdida en la generación; • pueden proveer estabilidad durante transitorios ya que amortigua las oscilaciones presentes en la línea de transmisión; • pueden amortiguar cambios bruscos de voltaje; • y finalmente, el sistema de almacenamiento en general es relativamente pequeño en tamaño en comparación con otros sistemas de almacenamiento y su ubicación no se ve limitada a algún área específica como es el caso de las hidroeléctricas. [10,4]. Figura 2. Tiempo de Descarga vs. Capacidad de almacenamiento en distintos sistemas de almacenamientos de energía. Figura modificada de la versión original , traducida al español. El primer SMES utilizado tanto para experimentación como para uso comercial fue diseñado por el Laboratorio Nacional Los Alamos (LANL, por sus siglas en inglés) y construido para la Bonnevile Power Company en 1982. Estuvo en uso por cinco años y fue desmantelado para investigación. Este proyecto tenía una capacidad energética de 30 MJ y fue utilizado para estabilizar el sistema de potencia ya que amortiguaba las oscilaciones presentes en una línea de transmisión de 1500 km de largo. Según LANL, el costo de construcción de un sistema de almacenamiento por superconducción se distribuye de la siguiente manera: • Bobina superconductora, 45%. • Estructura, 30%. • Mano de obra, 12%. • Convertidor, 8%. • Sistema de enfriamiento, 5%. El mayor reto que presenta esta tecnología es reducir el costo total del sistema. La Tabla 3 presenta una proyección de costo para distintos sistemas de almacenamiento de energía . Actualmente, el costo del sistema de almacenamiento SMES depende del costo de los superconductores. En el 2007, el costo del NbTi era de 1 $/ kAm, mientras que el Nb3Sn era de 1.50 $/kAm (Dólar/kilo ampere por metro). Afortunadamente, una característica inusual de este sistema es que el costo por unidad de energía almacenada (MJ o kW-hr) decrece a medida que la capacidad de almacenamiento aumenta, es por esto que este tipo de sistema es preferible para aplicaciones de gran tamaño como lo es el sistema de potencia, aunque también se están realizando estudios para su aplicación en vehículos . 32 Prisma T ecnológico | Vol. 4, n.° 1, edición 2013. Tabla 3. Proyección de costos para sistemas de almacenamiento de energía Sistema Tamaños Típicos MW $/kW $/kWh Ultra capacitores 1-10 300 3600 Volantes de Inercia 1-10 200-500 100-800 SMES 10-1000 300-1000 300-000 Aire Comprimido 50-1000 500-1000 10-15 Bombeo Hidroeléctrico 100-1000 600-1000 10-15 5. Conclusión SMES es una tecnología de almacenamiento de energía que tiene mucho potencial debido a su capacidad de almacenar grandes cantidades de energía y aún ser rentable en comparación con otros sistemas de almacenamiento. El mayor reto que presenta es reducir el costo total del sistema, pero avances en los sistemas de refrigeración criogénica y el desarrollo de mejores superconductores puede llevar a que su aplicación sea mucho más comercial. Referencias R. L. Evans, Fueling our Future: An Introduction to Sustainable Energy, 1era edición, Cambridge University Press, 2007, pp. 168. W. V. Hassenzahl, “Superconducting Magnetic Energy Storage,” IEEE Procee-dings, vol.71, No. 9, Septiembre, 1983. H.Y. Jung, A. Kim, J. Tamura, et. al, “A Study on the Operating Characteristics of SMES for Dispersed Power Generation System”, IEEE Trans. on Applied Super-conductivity, vol.19, No.3, Junio 2009, pp.2028-2031. W. Buckles and W. V. Hassenzahl, “Superconducting Magnetic Energy Storage,” IEEE Power Engineering Review, Mayo, 2000, pp.16-20. J. Bray, “Superconductors in Applications; Some Practical Aspects”, IEEE Trans. on Applied Superconductivity, vol.13, No.3, Junio 2009, pp.2533-2539. A. Ter-Gazarian, Energy Storage for Power Systems, 1st edition, Peter Peregri-nus Ltd., 1994. W. V. Hassenzahl, “Will Superconducting Magnetic Energy Storage be Used on Electric Utilities Systems?” IEEE trans. On Magnetics, vol.Mag-11, No. 2, Marzo, 1975, pp.482-488 S. Vasquez, S.M. Lukic, E. Galvan and J. Carrasco, Energy Storage Systems for Transport and Grid Applications, IEEE trans. On Industrial Electronics, vol.57, No.12, Diciembre 2010, pp.3881-3895 F. Farret and M. Godoy Simoes, “Integration of Alternative Sources of Energy”, IEEE Press/Wiley-Interscience, 2006, Hoboken New Jersey, 1era edición,pp. 290. L. Trevisani, A. Morandi, F. Negrini, P.L. Ribani and M. Fabbri, “Cryogenic Fuel-Cooled SMES for Hybrid Vehicle Application,” IEEE trans. On Applied Supercon-ductivity, vol. 19, No.3, Junio, 2009. Tecnología a fondo
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Solved: Square Numbers 1-20 10^2= 20^2= [Math] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Math Questions Question Square Numbers 1-20 10^2= 20^2= Gauth AI Solution 100%(1 rated) Answer $$10^{2}=100$$1 0 2=100 $$20^{2}=400$$2 0 2=400 Explanation Description: The image shows a table with the title "Square Numbers 1-20". The table has two columns, each listing the squares of numbers from 1 to 20. Explanation: To find the square of a number, you multiply the number by itself. For $$10^{2}$$1 0 2, we have 10 10 = 100. For $$20^{2}$$2 0 2, we have 20 20 = 400. Helpful Not Helpful Explain Simplify this solution Gauth AI Pro Back-to-School 3 Day Free Trial Limited offer! Enjoy unlimited answers for free. Join Gauth PLUS for $0 Previous questionNext question Related Solve the fallowing. 42= 112= 102= 202= 62= 4 Leslie baked 24 cupcakes Saturday. If of the cupcakes were chocolate, how many were NOT chacolate? 100% (5 rated) Work on all problems on the worksheet: 1-20 D 23 square root of 16 11 square root of s |2| square root of 101 3 square root of 36 4 square root of 64 13 square root of 125 14 square root of 50 J S square root of 80 6 square root of 30 15 square root of 175 16 square root of 28 8 square root of 18 17 square root of 45 18 square root of 72 9 square root of 32 10 square root of 12 19 square root of 20 20 square root of 150 1 square root of 75 square root of 64 16 square root of 28 4 100% (2 rated) Work on all problems on the worksheet: 1-20 n square root of 13 5 square root of 3 square root of 16 1 square root of s 2 square root of 108 overline Fe 40 square root of 64 3 square root of 36 13 square root of 125 14j square root of 50 9 square root of 80 6 square root of 30 5 square root of 175 16 square root of n 7 square root of s square root of 18 17 square root of 45 square root of 72 9 square root of 32 109 square root of 12 9 square root of 20 20 square root of 150 1 square root of 75 square root of 64 16 square root of 28 4 100% (2 rated) Tasha assembled a picture frame that is advertised as rectangular. The completed frame is 14 inches long and 10 inches wide. She measured the diagonal length across the frame as 20 inches. Which best explains why the frame cannot actually be rectangular? 142+102=202 14+10neq 20 14+102=202 142+1022=202 100% (4 rated) In a right triangle, if one acute angle is 45 ° , what is the measure of the other acute angle? 60 ° 90 ° 30 ° 45 ° 100% (1 rated) Part III: Substitute your results from Part II into the first equation. Solve to find the corresponding values of x. Show your work. 2 points Part IV: Write your solutions from Parts II and III as ordered pairs. 2 points __ and _ ' _ 100% (2 rated) How may different arrangements are there of the letters in The number of possible arrangements is MISSISSIPPI？ 100% (2 rated) Multiply and simplify the following. 3-i-4-9i -21-24i -21-23i -21+23i ⑤ 21-23i 100% (5 rated) The product of eight and seven when multiplied by F is less than the product of four and seven plus ten. a. 8+7F<4+7+10 b. 87F>47+10 C. 87F ≤ 47+10 d. 87F<47+10 100% (5 rated) Write the quotient in the form a+bi. 7-i/3-6i 7-i/3-6i =square Simplify your answer. Type your answer in the form a+bi . Use integers or fractions for any numbers in the expressio 100% (4 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App